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STRUCTURAL 

ENGINEERING 

PROBLEMS 

DEALING WITH FRAMES. 
WIND BRACING. RETAIN¬ 
ING WALLS. SHEET PILING. 

AND WAVE PRESSURE ON 
BREAKWATERS 


hy 

DAVID A. MOLITOB^ 

JB.C.E; (lE^Dr£iy.,yK£m.J)mSoc.C.E^ etc. 
Siruatit/xzl Engineer 


REPRODUCED BY THE BUREAU OF YARDS AND 
DOCKS. NAVY DEPARTMENT. WITH 
PERMISSION OF THE AUTHOR 

•193 7 - 



Copyright 1937 

by 

David A. Molitor. 


I- > 

^ .> 


MAR 30 1937 


©ClA A 232202 


T & 2-60 


PREFACE. 


The present treatise represents an effort to provide 
solutions for problems which have not received adequate 
consideration in our present day literature. It should 
serve a useful purpose to the practicing engineer and to 
students in structural engineering. 

The subjects dealt with are statically Indeterminate 
beams, colimms and rigid frames, especially those met with 
in modem buildings of steel and reinforced concrete 
construction. Wind bracing in tall buildings receives 
special consideration, while the chapters on retaining walls, 
sheet piling and wave pressure on breakwaters, offer a 
distinct contribution to engineering literature. 

In the opening chapters pertaining to statically 
indeterminate structures, the foxxndatlon is laid for a 
more comprehensive tinderstanding of structural problems, 
by developing methods for their diagnosis, followed by 
studies of principal systems, and the general laws govern¬ 
ing them, culminating in the selection of a method of stress 
analysis by Mohr’s work equation, which is practically 
universal in its application. 

The solution of rigid frame problems by Castigliano’s 
law, as generally advocated by American writers, involves 
the use of the differential and integral calculus. It is 
for this reason that the author advocates the use of 
Mohr’s work equation by means of ihich, and substitution 
formxilas (Table 26-A, here published for the first time), 
integrations are entirely eliminated. 

Hence the method here advocated invariably leads to 
a solution of any problem involving redundancy, avoiding 
the multiplicity of methods and the confusion which has 
been injected into our literature through the substitution 
of new names for old things. 

The subject of influence lines, so important in 
designing bridges with moving loads, and which was 
exhaustively treated in the author’s "Kinetic Theory of 
Engineering Structures" (now out of print) is entirely 
omitted, because problems in building construction usually 
involve only one position of live load for maximum effects, 
thus rendering their use superfluous for present needs. 

The volvane thus comprises the following chapters: 

Chapter 1, criteria for the.diagnosis of structural 
problems, not exhaustively treated elsewhere* 

Chapter 2, discusses the principal system in any 
structure 'Involving redundancy, and the choice in its 
selection. 





Chapter 3, presents the fundamental laws governing 
structures which every structural engineer should understand* 

Chapter 4, gives the general outline leading to the 
solution or all structural problems Involving redundancy* 

Chapter 5^ treats of deflection problems by area 
moment8 and Mohr*s work equation. 

Chapter 6, rounds out this portion of the treatise by 
presenting an extensive collection of problems all solved 
by Mohr*8 work equation, using substitution formulas 
Table 26-A 

Chapter 7 * on wind stresses in building frames, deals 
with approximate methods of designing wind systems for tall 
buildings, and detailing beam and column connections* 

Chapter 8 , on continuous concrete beams, should prove 
interesting and valuable in that the method advocated is 
simple and time saving, and leads to economical designs 
which have proven most satisfactory in practice. 

Chapter 9 , deals with earth pressure and retaining 
walls from a practical standpoint, giving everything that 
is useful and avoiding all that is unnecessary* 

Chapter 10 , presents problems relating te sheet 
pi ling and" r e ve tm en t s , treated in a comprehensive manner* 

Chapter 11 on wave pressure, sea walls and breakwaters, 
coverf a ihbjirdt about which much has been written without 
contributing much of value* The stability analysis here 
presented for the first time, is novel and original* 

In the words of the great Maeterlink, "It is to our 
humblest efforts that every useful, enduring achievement 
of this earth is due"* May this humble effort contribute 
something usefid to the chosen profession of the Author. 











STRUCTURAL ENGINEERING PROBLEMS 

CONTENTS. 


Cha-p, 1. Criteria for the Disunosls of Struotiaral Problems . 

Page. 

Art.. 1. Supports external to a structure._ i 

Art, £. Tests for redundaney of pin connected frames,_ 2 

Art. 3, Redundancy in rigid frames._ zj. 

Art. 4. Problems illustrating structural types.- 7 

Chap, 2. The Principal System of a Structure Involving 

Redundancy. 

Art. 5. The principal system of a structure,__ 8 

Art* 6 . Problems illustrating principal systems.- 9 

Chap* 3^ Fundamental Laws GoTerning Structures. 

Art* 7. Introductory._ 

Art* 8 . Law of elastic deformations._ 12 

Art. 9* Law of the summation of similar partial effects. 13 

Art* 10. Law of proportionality between cause and effect, 13 

Art. 11. Law of summation with redundants*- J 3 

Art. 12. General work equation. Clapeyron*s Law._ 15 

Art, 13. Law of virtual work. Lagrange.- 

Art. 14. Mohr^s work equations.- ,5 

Art* 16. Maxwell^s law,- 17 

Art, 16. Menabrea's law. theorem of least work.- )8 

Art. 17, Castigliano* 8 . law.- )9 

Cap, 4, Statioally Indeterminate Structures. 

Art, 18. Indeterminate frames by Mohr’s work equation.- 20 

Art* 19. Indeterminate solid web structures by Mohr’s work 

equation.- 21 

Art, 20. Indeterminate striiotures by Maxwell’s law.- 22 

Chap. 6 . Defleotions. 

Art. 21. The variety of deflection probleias.- 23 


I 


























Pacj e. 


Art* 2E. D«fl«otlona of beams by area moments*-24 

Art* £3* Grapkio solution of deflections by area moments*- Z5 

Art* £4* Slope of tbe elastic ourye by area moments*- 27 

Art* £5* Defleotiens of frames by Mokr's work eqxiation*- 2.1 

Art* £6* Befleotion of solid web struotures by Mohr*8 work 

equation*_30 

Chap* 6, Problems by Mohr*s Work Equation* 

Art* £T* ]>efleetion problems*-32 


Art* £6* The three*moment equation and its applications* 34 
Art* £9* Statioally indeterminate beamB and rigid frames* 40 


Chap* 7, Wind Stresses in Building Frames* 

Art* SO* The nature of the problem*-63 

Art* SI* T&e Tslidity of the assumptions* 63 

Art* S£* Stress analysis of a giren design*-65 

Art* SS* Approximate method of design*_65 

Art« S4* More accurate solution for bottom stories*-69 

Art* S6* Comparison of results due to seyeral methods*-- 71 

Art* S6* Beam and column wind oonneotions*_73 

Art* S7* Detoils for beam and column wind connections*- 75 

Art* S8* Wind pressure*_- 75 


Concrete Beams. 

Art* S9* Description of method*__ 80 

Art* 40* Example*_ 53 

Art* 41* rmstification of the method.__g 5 

Chap* 9* Retaining Walls* 

Art* 4£* Earth pressure on the back of a wall*_ 

Art* 4S« Earth pressure on walls without surcharge*_ 9 ^ 

Art* 44* Craphieal solution* wall without sureharge*_ 9 ) 


II 




















Art. 45* CrTaphioal solution» walls with surohargo._ 93 

Art* 46* Praotioal wall design data*_94 

Art* 47. Probleas*_ \oo 

Art* 46* Arches under embanlments_ 104 

Chap* 10, Shaat Piling probleaa. 

49* Sheet piling in lewel earth with horisontal foroe at 

Top *- 112 . 

Art* 50* Sheet piling baok'filled with earth*-113 

Art* 61. The effeot of reliOTing platforms._114 

Art* 5£* Sheet piling backfilled and with baok anohors._(|5 

Art* 5S* The ooabined action of earth and water on sheet 

piling.- 116 

Art* 54. Resistance deweloped by back anchors*-122 


Art* 66 * Length of anchor rods and location of anchor plates.)Z3 
Chap* 11* Ware pressure, Sea Walla, Breaks Waters .** 

Art* 66 * On the nature olthe problem-J25 

Art. 67. Height of wsTes in terns of wind Tsloolty and fetch.126 
Art* 68 * Length of ways in terns of height and wind Telocity. 127 
Art. 69. Lyname properties of wayes*- 129 

Art. 60* Ways pressure obseryationa. Lakes Superior and 

Ontario*-131 

Art* 61. Stability of breakwater orlbs subjeeted to ways foroei41 

Art. 6£* Conoludiag renarks*- 146 

(Wiap* IS. ELsTstsd Steel Qrliadrioal Tanks.-X47 


Note: Chapter* Jl was printed in Trans-Am. Soc. C.E. Vol. )oo(i935) 


















CHAR I-CRITERIA FOR THS DIAGNOSIS OF STRUCTURAL FROBLblMS, 


Structural Suffici e ncy and Rediindancy. 


The purpose of structures is to sustain loads and subsequently 
transfer them to the supporting earth or foundation. 


Structures may consist of rigid frames, pin connected frames or 
solid web beams, or of combinations of any of those typos, 

kVhen the members of a structure occupy two or more pianos we 
speak of it as a framework in space. All of the structures which are to 
bo considered here will consist of members in a single plane. 


ART. 1, 


Supports external to a structure. 


1 . Siinple contact between frictionless surfaces offers 
resistance in one direction only. This is exemplified in a roller 
bearing or hinged pendulum, and supplies one reaction condition, pi^. i-A 

2. A hinged .joint or pin end connection to the earth is 
capable of resisting motion vertically and horizontally, yet permitting 
of rotation. Sueh a support affords two reaction conditions. Fig. I-B. 


5. A rigid support resists motion In all directiens and hence 
involves three reaction conditions. Fig. I-C. 




These may be represented by two forces and a moment or by throe forces. 
.Vhen the three forces are parallel the case reverts to one reaction 
condition, and when the throe forces intersect in a commom point the 
case reverts to two reaction conditions. 

The three types of supports, movable# hinged and fixed, thus 
Involve respectively one# two and three reaction conditions, and each 
reaction condition may be represented by a short link or member capable 
of resisting tension or compression. 


1 


























No structure, however simple, can be supported on the ground by 
less than three reaction conditions, this being the absolute minimum 
requirement for stability. When several frames are combined into a 
composite structure as for a three hinged arch or girder on more than two 
supports, or a cantilever on more than one suppott, then the number of 
necessary reaction conditions to insure stability will be i- 

^ r at * e+2 

where e is the number of elements or simple frames in the composite 
structure, 


Hence S.r-e-2 s n' represents the number of external 
redundant conditions for any given structure composed of e elements. 

The conditions for static equilibrium applied to any structure 
as a whole whether simple or composite, are represented by the equations 

=5 0 0 = 0 

where the stanmation includes all the externally applied loads Including 
the reactions. 

These same condition equations apply equally to all forces and 
strcaaes in members active about any pin point of any structure. 

c 

When the external forces are not thus balanced among themselves 
and with the internal stresses in its members, then the structure is 
unstable or statically insufficient. 

A structure is therefore externally determinate when the three 
condition equations suffice to determcne the reactions, which will also 
be tne case when :tr — e + 2 . 

A framed structure ( pin connected ) is internally determinate 
when its members are arranged to form triangles in such a manner that the 
successive removal of pairs of members uniting in a point, finally reduces 
the strueture to a single triangle. The triangle is^therefore, the 
primary element of all determinate pin connected frames. 

A structure is in every sense determinate when all reactions 
and stresses can be expressed purely as functions of the externally applied 
loads, unaffected by all temperature changes or small reaction displacements. 


ART. 2. 


TESTS FOR REDUNDANCY OF PIN CONNECTED FRAMES. 


Let 


P 

m 

n 

n* 


. 1 * 


e 


number of pin or panel points of any given frame, 
number of members in the frame, 
total niimber of redundant conditions, 
number of external redundants. 

" " internal " 

number of elements or simple frames in the structure, 
number of actual reaction conditions. 


2 







Then for any statically determinate structure, 

2 p — m + .(2A) 

when this condition is satisfied, the frame is always statically 
determinate but not always stable, for it may be dynamically insufficient 
as will be explained later. 


When 2 p > m + 2 r .(2B) 

then the structure is not in stable equilibrium and is statically insufficienjb. 

When 2 p <. m + ^ r .(2C) 

then the structure involves redundant conditions, the number of which is 
given by the equation 

n m + :gr-2p .(2D) 

wherein a negative n would Indicate static insufficiency. 


In all the equations (2A) to (2D) the total number of z*eaction 
conditions for any simple or composite structure, if statically determinate, 
is t - 


^ r e + 2 


(2E) 


The number of eatemal redimdant conditions for any given structure 
eompoeed of e elements is t- 

n*-^r-e-2 .(2F) 

and hence the number o^ internal redundant eonditions must be t- 

n** sc m - n* ...(20) 


It is thus seen that equations (2D) and (2F) will answer every question 
relating to stability euid redundancy of any simple or composite etnictture 
composed of pin connected trieuigles. The same equations also apply to 
solid web structures where the restrained ends constitute supports to the 
earth. This excludes the class of structures known as rigid, frames 
which will now be discussed. 

It aheuld be noted that temperature stresses are produced only 
when there is external redundancy. 


3 










ART> ^ 


R3DUNDANCY IN RIGID FRAMES 


In treating rigid frames it is agaiii necessary to distinguish 
between external and internal redundancy. This is apparent from the 
fact that any rigid frame which does not close on itself may involve 
external redundancy due to the manner in which it is supported. 

Thus two and three sided frames, also continuous and restrained 
beams are all externally redundant, while multiple story bents, four sided 
frames and many other types of rigid frames cannot be converted into a 
statically determinate structure simply by rearranging the supports. 

These involve internal redundancy in addition to the external. 

The characteristic of a rigid joint as distinguished from a pin 
joint consists of a conventional knee brace which prevents relative 
rotation of one member with respect to another. Thus a pin joint will 
resist horizontal and vertical forces only, while a rigid joint will resist 
rotation as well. 

As an illustration consider a four sided rigid frame Fig, jJA , 
on statically determinate supports. 



Fig.3A 





Fig- 3‘B. 


If all ftawr knee braces were removed. Fig. 5B, the frame would collapse, 
independently of how it*was supported. By replacing one of the knee 
braces, or by inserting a diagonal member with pin end connections, 
stability would be again established so that we could dispense with at 
moat three braces, and thus decide that the frame had three internal 
redundants. 


Another illustration of a multiple three sided frame. Fig JC, 
with poets hinged at the bottom, shows all the rigid joints replaced 
with conventional knee braces. 


A 

3 < 

2 

D 

’ 

yTTTrrn 

ynM. 

7 

7 V 

^rrTTTTrrrrTTft' 

fTrTmrrrrrr^ 

7 V 


Fig. 3C. 


4 













In this frame wo could dispense with seven knee braces, retaining 
merely one to prevent collapse of the structure. Hence we are here confronted 
with seven internal redundants and no external reduntants. 


iVe will now consider a frame similar to Fig. ^C, but in which all the 
posts are rigidly connected to the earth by a brace at the foot of each post. 
Only one brace per post is necessary to produce a fixed end because the earth 
acta as a continuous member between posts. Fig. 5D shows this frame with all 
posts fixed at the bottom. 


B 


. 


A 


Fig. 3-D. 


We could now remove three braces from each bay or a total of twelve, retaining 
only one brace to avoid collapse. Hence we now have a total of twelve redundants 
of which at least four are external if we remove all of the top braces. Wo 
would have five external and seven internal redundants if all bottom braces were 
removed, retaining one top brace. 

Let us consider one more case of a multiple story bent as shown in Fig. 52 

where all members are rigidly connected to 
each other and the bottom ends of the posts 
are fixed to the earth, »e now require 
54 conventional braces to exemplify all the 
rigid end connections involved in the frame, 
and 51 of these could be removed .vithout 
causing collapse provided the remaining 
three are suitably located, as for instance 
those marked 1, or the three marked 2, 5» 
and 4, or in short retaining one brace in 
each story. 

Fig. 3-E. 

lie may nov; foi-mulato a rule for the correct diagnosis of any rigid frame, 
based on the number m of rigid knees emd the number s of stories in the frame. 
The total number of redundants will be:- 

n s m - s .(5A) 

It is usually quite easy to count off the number n’ of external redundants 
involved, thus leaving n - n* internal redundants. 

The method of successively removing the conventional knee braces from 
any rigid frame finally leads to a principal system as in the case of pin 
connected structures. This principal system forms the statically determinate 

portion of the rigid frame from which nothing further can be removed without 
causing collapse. 

The introduction of the conventional knee brace offers a simple expedient 
in the diagnosis of rigid frame problems. One feature, however, which may be 
overlooked is the case of a rigid joint combi'ned with an actual knee brace as is 
often employed in bridge portals of riveted steel construction. 


r ,—^ 

S 

?-< 

313^ Story, 

’-7! 


7 V 

Story. 
S <1 


T, ^ 

i- -4 

story. 

/ V 

4 


5 























Wo then have the actual knee aa an extra redundant in excess of the rigid joint 
with conventional knee only. 

It should now be easy to contemplate the difficulties attending the 
analysis of many of our everyday problems in structural design^ which are, 
however, of a purely mathematical nature, and may be solved if tine permits and 
the endurance limit is not exceeded. 

There are practical considerations and limitations which preclude the 
advisability of accurate analyses of most of these involved problems for the 
very simple reason that many uncertainties exist which greatly outweigh or 
overshadow relatively large approximations which may generally be permissible 
in expediting our work, 

A tev of the commonly neglected elements v/hich affect ?ur structures, 
sometimes on the side of safety and at other times adversely or even dangerously, 
may deserve mention here. 

For instance,there is no such thing as a frictionless pin connection, 
yet this is commonly assumed in our computations of pin connected frames. On the 
other hand, there is no such thing as complete restraint which is a physical 
impossibility, yet we regard so-called riveted connections and concrete frames 
as rigid. The truth of the matter is evidently some intermediate status of 
fixity in both cases, and may differ widely from our assumptions in calculating 
stresses. 


In reinforced concrete structures we encounter a variety of uncertainties 
which defeat all efforts tov/ard accuracy in design. Thus in buildings the 
unavoidable restraint between floor-slabs and beams make it impossible to estimate 
the efiective cross section of T beams. Also, the restraint between beams, 
girders and columns introduces uncertainties which must render our calculations 
grossly approximate, 

Elven in steel frames a considerable restraint is always developed by the 
standard end connections to other beams and columns, and yet we base our calculations 
on simple beam assumptions. 

Under such circumstances we are amply justified in adopting approximate 
methods of design so long as our errors are on the side of safety instead of 
bordering on recklessness. 

With these facts in mind we can usually simplify the solution of complex 
problems by neglecting those factors which ahe of minor importance, but add 
greatly to the labors of the designer. However, to exercise this criterion with 
prudence, we must acquire a ra'’turity of judgement v/hich canbe attained only in 
the workshop of experience. 


6 



ART4-. PROBLEMS ILLUSTRATING- STRUCTURAL TTPES. 





“•1 

uo. 

£r 

m 

p 

e 

Remarks ] 

H 


m 

P 

e 

Remarks- 

Pig 

i.r 

m 

P 

e 

Remarks. 

1 

4 

1 

2 

1 

i ext. redundan^ 

r’ 

6 

19 

11 


3 exi.redundanh 

13 

s 

45 

24 

1 

2 ext.redundents 

2 

6 

1 

2 

1 

3 .. 

s 

3 

9 

6 

1 

infin&e stress. 

14 

8 

3 

4 

3 

3 .. 

3 

6 

1 

2 

1 

3 - 

9 

6 

10 

7 

2 

4ext.redundanf$ 

15 

3 

4 

- 

- 

3 int. ** 

4 

3 

s 

5 

1 

1 inf. 

10 

3 

17 

10 

1 

Unstoble 

16 

4 

7 

- 

— 

6 i«t. « 

5 

4 

lo 

7 

2 

Sfaf.detcrm. [ 

11 

3 

36 

19 

1 

1 1 ’nt.red- 

17 

6 

4 

- 

— 

3 ext- 

6 

4 


13 

1 

icxf. redund. | 

1 12 

3 

37 

20 

1 

Sfat. deUrm. 








7 


















































































































CHAR2 - The Principal Syatem of a Structure Involving Redundancy . 


Art. 5 . The crrtcria for recognizing the otatically detenninate portion or 
principal syatem of any structure involving redundancy, were discussed in the 
previous chapter. 'He will now investigate the nature and form which a 
principal system may assume for any redundant conditions as may be found to 
exist in any given structure, 

A principal system may be defined as any statically deteminate 
portion of a structure from which all external and internal redundant conditions 
have been removed. The conception of a principal system affords a means for the 
comprehensive and universally applicable method of analysis of all problems 
involving redundancy. It was first introduced by Prof,Otto Mohr in 187^, and 
constitutes the basis for his well knovm v^ork equations. 

Having ascertained the number of redundants involved in a given example, 
by applying our criteria, we then proceed to remove them so as to arrive at some 
suitable principal system. We then reapply these several assumed redundants 
along with the externally applied loading, to ascertain the combined effect of 
all on whatever principal system was selected. 

The principal system being statically determinate, permits of finding 
all functions pertaining thereto by the ordinary methods of statics, for the 
severally assumed redundants and the actual externally applied loading. 

Hence, if we knov/ how to find reactions, moments, she'rs, stresses 
and deflections for any statically determinf^te stiructure, then we possess the 
tools for solving redundancy problems by applying Mohr's work equation to the 
principal system. This will be dealt with in Chapter four. 

Our next effort will be directed to the choice of a principal system 
for any ■ iven problem, for much depends on making a judicious choice in order 
that our subsequent labors may be minimised. 

In all our investigations of statically indeterminate structures the 
redundant conditions, whether reaction forces, moments, shears or stresses in 
■members, will always be designated by Xa>^b> » etc, implying that they 
represent those unknowns which must first be evaluated from the elastic 
properties of t/ie materials of construction, and the geometric relations 
presented by the structural dimensions. The actual evaluation of those 
redundnats X is accomplished by means of elasticity equations according to 
Mohr, of which there will always be as many as the number of redundants. 

That is,one work equation can be written for each redundant thus furnishing 
as many equations as there are unknowns X. With the X's evaluated, the 
rernaindar of the analysis reverts to the ordinary methods of statics as 
applied to the principal system, wherein the X's are treated like known external 
forces. 


The particular reactions or members best suited to represent the 
redundant conditions designated byXA,X|,,Xc,otc., are those which reduce the 
given structure to the simplest possible principal sj^stem. 

Any member or reaction of an indeterminate structure may be removed to 
produce the principal system so long as the latter still remains a stable 
determinate structure. 


8 








However, the rule should be to select a principal system of the 
simplest possible form, always avoiding composite structures such as a three- 
hinged arch or a cantilever system ,wHtfHgVgfa simple beam, cantilever, or truss 
could as well be chosen. 


Solid web structures should always be transformed into externally 
determinate beans by assigning the redundant conditions to the supports. 

Framed structures. If externally indeterminate, should always be so 
transformed as to remove the ewternal conditions. The only exception to this 
rule might be a continuous girder wherein a top chord member over each inter¬ 
mediate pier might be treated as a redundant member. 


when the redundant conditions are internal, then the only way of 
deriving the principal system is to remove such redundant members and replace 
each one by two equal and opposite forces X, 

Art. 6. Problems illustrating Principal Systems. 

A number of problems v/ill be presented here with a view of showing the 
various possibilities in the choice of a principal system, and the manner in 
which any redundants may be disposed of by means of conventional loadings^ 
reapplied to the principal system. 


Problem 6 A. Simple beam fixed at one end and supported on a roller bearing 
at the other end. 




iL 


Given case. 




r 4 , m = 1, jp_ 


e = 1 


4 — . - ■ — ■' ■ — 

^ -:-:-:- 

^ Principal system. 

^ Condition X«o. 


I 


Conventionol Loading 
Condition X^«l. 

F,i.(a). 




n r ir + m - 2 p a 1 

n'=:gr-e-2sl 

One external redundant. 

a. Principal system a cantilever , obtained 
by removing the roller bearing and 
calling the end reaction the redundantX*. 

6ondition X ■ 0. 

The conventional loadingapplied 
to the principal system to the exclusion 
of all other loads. Condition I. 

When Xft is found for the actual case of external loading, than every other feature 
of the problem is solvable by the methods known to statics. 

^ Ip 

Principal system a simple beam obtained 
by converting the fixed end support into 
a pin end support with an external 
redundant moment. Condition X ■ 0, 

__ The conventional loading I is applied 

f . - .. .. — A to the principal system and pictures the 

C,nv.r.fionaU^.n,.-Y- conditlL X,= I. 

on I ion ^ . When is found for the actual case of 

external loading, the problem is again 
solvable by statics eind will lead to 
answers which are numeriealljjC'^WtoAtiial with the cantilever solution. There is 
not much choice between these two solutions as they are equally simple. 




Principal system. 
Condition X«0. ' 




9 

























Problem 6 B 


Beeun on three aupporta 


I Li 


I IIJ 


t, —t, ~fc a|.-- t. 

I Given Case. I Principoi System. ConditlonXv-sl. 


Principal System. 
Solufjon( 3 .] A simple beom. 

F F 

2r.4,m=l. p.2, e.i 1 V 

1 rerfunclanf. "T" _ • • T 

• • HV Principal System. 

ns4-l-2*s|«>ct. SolutlOn{^bj z-simple beams. 

F f f 


ConditlonX^»l. 
Center Support redundant. 

- J. - - - ^ 

I Condition X-j,®I. I 
Center momentredundant. 


f 


. r V Principal System 
Solutjo n(cj simple beam andcontilever 


r=^ 


Condition X^fsl 
ITndSuoDort redundant 


Problem 6 C , Two hinged arch. 



a6r»4 , ms|, ps2, ea I. 

n « ^■r I —Zit'Z* I redundant. 

0 

ns 4—1—2 =1 e«t. redundant. 


Principal System. 
A 3-hiii<)ed arch 


Condition X^s I. 
Center moment redund. 


Problem 6 D. Fixed solid v.eb arch rib. 



acrs4 , m»i, ps2 ,€»!. 
n » fe-*-1 —2*2 « 3 redundanfs. 
n*3» 6-1—2 =3 external. 



Solution (a.). 



Applied io each Caniilever in 
oppoiite sente far each arm. 


10 






















































Problem 6 D, continued 


r 


P 

--- 

P 


P 








A 

' 

r 





B 

Principal System 
simple Cantilever. 

3 redundentS; Moment X^ 

Principal System 
simple beom. 

3 redundents; MomentX^ 



Her. thrust Xe 
V«rt. React. 

Solution ("b) 


Moment 
Hor. thrust 


Solution (C) 




Xe p 


B 


Principal System 
3-Hinged arch. 

3 re<4undents; Moment 

MomentXi,, 
Moment X^ ^ 

Solution (d). 


Case e glTos preferable solution. 


Problem 6 E. Fixed Beom and Column with Cantilever. 


airsO, ntmt , ps2, «s|. 
n 3 6’et>2K2n 3 red undents, 
n'm e-l - 2 ■ a e«t. redundnnfs. 



Principal System 
simple Cenfilercr. 

3 redundentsjHor.thrustX* 
Vert.Rceef.X;^ 
Moment X^ 

Solution (a) 



Principal System 
simple beam. 

S redundents; Moment X^^ 
Moment X|^ 
Hor.thrustX^ 

Solution Cb) 



Principal System 
S'Hin^ed arch. 

3 redundents; Moment . 

MomentX^. 
Moment . 

Solution (c). 


Case c gives preferable solution. 


11 





































Chapter Fundamental Laws 


Governing Frames and Solid Web Structures, both statically determinate 

and indeterminate. 


Art. 7. Introductory . The laws of statics suffice for the analysis of all statically 
determinate structures in so 'far as stresses are concerned, euid the elastic 
properties of the material are not involved. This method of treatment 
tacitly assumes that the structure is at rest. However, the moment when 
externally applied loads begin to act on a structure it ceases to be at rest, 
and must undergo elastic deformation so that we really have no license to 
regard an elastic structure as statically at rest^nly when it is neither stressed 
nor distorted. 

In abroader sense we must regard all structures as mechanical 
contrivances subject to some elastic motion, instead of inelastic bodies at 
rest. It is true that this motion prevails only while a change in the 
elastic balance is taking place, as when loads are added to, or removed from, 
the structure, but this may apply to any mechanism of interrupted activity. 

Therefore, vfnile wo speak of a structure,as in static equilibrium, 
v/e aety also speak of it as in a state of dynmanic equilibrium, u state which 
the structure assumes in the instant that the super-imposed loads do not 
produce any further deformations. The some would be true when i'll loads are 
entirely removed, in which case the dynamic equilibrium returns to the special 
case of static equilibrium. This is merely a broader view point involving 
the principles of work, embracing at once all of the conditions as they really 
exist in a structure sustaining loads. 

For the same structure, the magnitude of the deformation is a direct 
function of the applied loads, the unit stresses which they produce, and the 
elastic properties of the materials employed. 

The stiffness.or rigidity of a given structure, is however, independent 
of the magnitudes of the unit stresses, but depends entirely on the ability of 
the structure to resist stress, and this in turn is a function of the elasticity 
of the material and the geometric shape of the structure. 

With these concepts in mind, the general.laws pertaining to the 
comprehensive analysis of structures and their behavior under loads \;ill now be 
considered in brief review even though not aH of them will necessarily be applied 
to the solution of problems. 

The v.'ork of defomation constitutes the basis for the solution of all 
those problems v;hich involve the elastic properties of tlie material and v.hich are 
not susceptible to analysis by the methods of pure statics. 

Art.8 Lav/ of elastic deformations 

Let Sj* total stress in any member of a frame from any cause designating tension by-f 
tm length of this member when S«o. 

change in length C due to stress S, + for elongation. 

Fa cross-section of this member, uniform over length t . 
t= a uniform change in temperature in degrees + for rise. 
e= coefficient of linear exp-^nsion per degree temperature. 

Es liodulus of elasticity = unit stress divided by unit strain. 

S/F = unit stress in the member 
s ^ = the extensibility. 

unit strain. 


12 











Then according to Hook's lav:^ 


E s ^ ^ 

from which s -t ttt .(£a) 

EF ^ 

'N'aere lit represents the added change in length due to temperature. 


T’.'.is equation represents the elastic deformation for any member of any 
frsone amd is a fundamental elasticity condition. 

Art 9 Law of the summation of similar p- rr.iel ef ects . 

Every set of simultaneous causes or conditions, such as a load P, 
temperature rise t, euid reaction displacement acting on a given structure, 

produces a parti-^l stress S' which contributes in producing an ultimate total 
value S due to all partial values or effects resulting from the rjispective sets 
of independent causes or conditions. 


This is expressed algebraically by 

S s 6% S etc.(9A) 

and signifies that each effect, such as a stress, v/het ier due t o loads, tem. eratnre, 
or abutment disclacements, may be ascertained or investigated by itself, and the 
sum total effect S will then be the sum of the several similar partial ef".ects. 

This lav^ is fundamental to the analysis of all structures Involving 
redundancy. 


Art 10 Lav/ ~>t propo rt ionality betv/een cause and effect , 

Equation (9A) being true for any set of effects, v.ould remain true for 
any multiple of these effects. Thus if a set of loads P produces .stresses S 
in the members of a given frame,then another set of parallel loads ZP acting 

at the same points, v/ould produce stresses 2S in the sa.uo me.Tibors. Or, if 

a single load unity, acting on a point m of any frame produces reactions R, , 
stresses S,, and deflections 5, » then a load parallel to the unit load and 

acting on the same point m will produce reactions > stresses Sal^S, , 

and deflections $uf*S , 

IPI I 

.'.rt 11 Lav/ of sucrretion v/ith radundante. 


This law expresses the SLuacation of 'll partial effects due respectively 
to the applied loads, the several redundants, and to temperature c .anges, v.i.on 
acti^ig simultaneously on the principal sy.'tem of any pin connected frai.ie or solid 
v.'eb structure. The sum total effect sought inay be t!-;© stress in any member, the 
reaction at any support, or the beading moment about any point of the structure, 
and the partial effects contributing to any one total must necessarily be similar 
in kind or character. 

The strnicture under consideration must be so constituted that t..e removal 
of all the redundant conditions will produce a statically .determinate structure with 
the necessary supports. The statically determinate structure thus re ;• ir.ing i.ill 
always be called the principal system . The redundants, when applied to the principal 
system, v/ill always be designated as unimowns , etc, a id when singly 

applied to the principal system f.ie several cases of loading will be spoken of as 
conventional loadings . 


13 














When the redundants X (knovm or unknovm) are again applied to the 
principal system, together with the other external loads P, then the resulting 
effect on every portion of the principal system must be identical with th^ effect 
v/hich v/ould have been produced by the original loading of the given indeterminate 
structure. This will also be true of deformations. 

The following definitions of terms will be strictly adhered to in all 
succeeding discussions. 

Lot Sa the stress in any member of a principal system, 

S a the stress in this member due to loads P v/hen the several redundants 
X^and t are all r.ero, to be known as Condition X = 0 

Scathe stress in this member of the principal system when no load other 
than I is active. Condition I. 

Scathe same when no load other thsjiXjja I is active on the principal system. 
Condition |. 

Scathe same for Condition X^sl. 

the same for a uniform change in temperature. 

defined like the Ss with like subscripts, but "represent 

reactions. 

W*,Mj^.M^etc, are moments defined like the Ss with like subscripts. 

^*>X^, etc, are the redundants, and may be stresses in members, reactions, 
moments, etc, as may be assigned in reducing fne given case to a principal system. 

» «bc, are the displacements in the points of application of the 
redundants, for effects due to loads P and redundants X. 

The stress S in any member of a frame involving redundancy, is a linear 
function of the loads P, etc, all treated as external forces applied to the 

principal system. This follows because all conditions of equilibrium are represented 
by linear equations whether dealing with stresses, moments, or reactions. 


Now, according to the lav/ of the summation of similar partial effects 
expressed by equation (9A) the general equation for stress in any member of a frame 
involving redundancy, would have the form 

S — Sfl ■* dan ■” ^Vx “* ®> 

wherein S^^sthe stress in this member due to X,^ acting alone on the principal 
system, and a the stress in this member due to X^ acting alone on the 
principal system, etc. 

The negative signs applied to the stresses produced by the redundants 
indicate that t’.iese stress contributions are always of opposite sign to the stress 
contributed by external loads P . This follows because the increment of work 
performed by the redundants X is always negative with respect to the work performed 
by the stress S., because the redundants are classed with the external forces but 
actually assist in carrying them. 


Also,from the law of proportionality if follows that Sax*®j > 

•to. Hence for the stress in any member of the principal system 

S ~ Sq — S.^Xa'* S'bX'i,— S-j;----- — —) 

also for any reaction of the principal system )...(11A) 

R ~ ““ RflXa *” R^Xi, — chc. + R.j-__j 

and for any moment on the principal system j 

+ Mt_ I 


14 





In these equations the quantities S*, R»,and are all linear functions 
of the externally applied loads P, while all the stresses, reactions, and moments 
bearing subscripts a, b, c, etc, are constants due to conventional loadings and 
are absolutely independent of all external loads f and X applied to the principal 
system. 

These equations are fundamental in the analysis of all structures involving 
redundancy. 

Art, 12, General Work Equation, Clapeyron*s Law, 1853» 

The work of deformation of any structure sustaining loads, constitutes the 
basis for the solution of all those problems which involve the elastic properties of 
the materials and which are not susceptible to analysis by the methods of pure statics, 

A loaded frame is a machine in equilibrium and within the limits of 
proportional elasticity its deformation varies directly with the magnitude of the 
superimposed loads. 

The question of deformation does not enter into statics and hence a general 
and comprehensive treatment of the elastic frame must necessarily involve the principles 
of mechanics. 


A given frame with definite loading and supports constitutes a system in 
external equilibrium. The frame luidergoes certain deformations v^hich steadily 
increase in direct proportion with the internal stresses created in the members by 
the external forces as they are gradually applied. The final deformation will occur 
in the instant when the external forces are exactly balanced by the internal stresses* 
when a condition of static equilibrium is attained. 

During the process of deformation, the applied loads travel through distances 
which are the displacements of their points of application, producing the positive 
work of deformation of the external forces. 

The internal stresses in the members must accomodate themselves to the 
deformed condition of the frame and in resisting this action must produce negative 
work of deformation. 


In the instant \7hen static equilibrium is established between the loads 
and the stresses, the positive and negative work of deformation, produced in the 
some time interval, must exactly balance, according to the doctrine of the conservation 
of energy, 

A positive amount of v<ork is always produced when a forcg and its displace¬ 
ment act in the same direction. 


The product ofrepresents the actual work produced by a force gradually 
applied and increasing from its initial zero value to a certain end value P, thus 
exerting only its average intensity during the entire time of traversing the path 
to perform this work. Hence, for all the external forces P acting on any frame, the 
total positive external work of deformation T/ould be : - 


= PS. 

Similarly ^ represents the actual internal work resisted by any member 
subjected to a gradually increasing stress of eni value S, with an average intensity 
S/2 during the entire process while producing a change in its length of aC. 

Hence the stresses in all members in any frame will represent a total negative v/ork 
of deformation of »- 

ss -y S 5 A C . 

As the applied work must enual the work overcome, therefore, for any frame. 





<■( 

er 


(12A) 


which is Clapeyron's law and may be stated as follows t- 


15 









For any fre.me of constant temperature, and noted on by loads which are 
gradually applied, the actual .vork produced durin,<^ deformat log is independent of 

the manner in '.hich these loads are created and is al..a^'^s ''.alf as great as the 

— - - -— _ _ - - __t/_ — ^ - - - - — 

work otlierv.ise produced by forces retaining their full end v ines during t'ne entire 

act of defonuation. 


By regardin' any elastic body as composed of an infinite multiplicity of 
members, it is readily seen that Clapeyron's law^^**-ii?ally to fra '.es and solid wjb elastic 
structures v/hen properly supported. 

Cases involving dynamic impact would imply a certain amount of Icinnti'' '•;.ior'/ 
in excess of the negative work of defoxanation; that is, the ap’lied forces h-.v? some 
initial value, greater than zero, at a time when the stresses are still Z'ro. 

Hence e-uation (12A) does not apply to effects of falling boli'jr. 

Art, 1$, Lay/ of Virtual '.iork, 1788» Lagrange, 


This la.7 was first enunciated by Galilei and Stevin and later by Bernoulli, 
but Lagrange first reduced it to an algebraic expression which permits of a more 
varied application than Clapeyron*s law. The following derivation may be of intjrest. 


A.ll forces or stresses, acting on a pin point of any structure in e-uilibrium, 
ill have components parallel to '■ny axis, and tlxe sum of such, a set of “arallel 
components must be zero. Calling cC the angle v;hich any force or stress ^ makes .. ith 
the axis chosen, then for all such meeting in one point, 

S. CL Cos oC a O 

Mow if a displacement a , parallel to' th' axis, be srbitraril*/ assigned to 
this pin point, and assuming equilibrium to continue, then the product of a v.iVn 
*■16 sura of the components must still be zero. T'nis is ecuivalent to multiplying’ 
bot'n sides of the a’oove e'viatioji hv a to obtain s- 

S (acos<ac)A a o 

But the displace nents ACos^sa are the projections of tVie displaci-nent A 
on the directions of the several forces, hence »- 




(15A) 


•wherein the forces Q are in every sense independent of the displacements S , which 
latter may be any possible displacements resulting from some state of e uilibrium 
duo to another set of forces P in the same complex of membera. 


Hence for any point, frame, or body acted on by loads Q in an established 
state of equilibrium, the sum total work performed by these forces,in movin' over 
any small arbitrary bi^ possible displacements B , must al'ways be e-ual to zero. 

Art, 1*4 . Mohr’s V<ork Equations. 187^, 

Troceeding from the law of vFrtual -./ork as applied to two cases of loading, 
once for arbitrary loads Q producing stresses in equilibrium, and once for 
displacements 6 and changes in the lengths of members a 6 due to loads P in 
equilibrium, we can write for any frame »- 

and 32. So.A Cso^ and hence hj subtraction : — 

^ Qi -£ S ./.<»0 .. (14a) 

which represents a condition of olastlc equilibrium as distinguished from static 
equilibrium. 


16 
















Add a+the end of Art.'^.p. 16 


16 a 













1 



This lav.' expresses the equality between the external and internal 
work of deformation for real displacements and arbitrary cases of loading , 

provided equilibrium exists . 


By allowing all the arbitrary loads Q in equation ( 14 a ) to vanish, 
and substituting therefor a single load unity at a point m, producing stresses 
So. (and reactions in the members of a frame, then the new work equation 

becomes i- 

|.S„+ SR.ar = 2 S.a£ .(I'tB) 

whareln Sm is the displacement of the point of application of the unit load 
at m in the direction of this load, and a( successively for each member, is 
the change in length of such member, due to loads P producing a set of stresses 
S in the members, and abutment displacements a r at the supports. 

Since when temperature effects are also included, then 

equation (143) becomes in Its most comprehensive form 

l.S„ ... (I4c) 


Omitting temperature effects and abutment displacements, we have the 


usual form 


i.S^-ss , 


,(14D) 


for Mohr’s work eouation. 


The unit loading producing stresses S,,. and reactions v/ill always 
bo called the conventional loading and may be a single force unity,or a moment 
equal to unity, applied at a point m. 

For solid web structures equation (14 d) must be transformed to include 
the effects due to direct stress and bending, and then becomes 

N^djc , .(1^2) 

v/herein N - the direct axial stress due to loads P 

M s the bending moment at any point due to loads P 

N<^=the direct axial stress due to a conventional unit loading at m 
sthe bending moment at any point due to the conventional unit 
^ loading at m . 

S = displacement of point m in the direction of the unit conventional 

load, and due to actual loads P. 


For a single load -5=. ^the deflection of the point m 
becomes by equations (14 d) (14s) 




EF 



O’*! 



=— f 

^M^dx 



Pm ef 

Pm«y. 


Art 15. 

Maxwell's Law. 

“18^. 



in the direction of P,^ 

.(14f) 


The theorem generally known as the law of reciprocal displacements 
establishes the mutual relation between the elastic displacements of any two 
points of any structure, whenever these displacements result from simultaneous 
conditions of loading) provided that the members remain unchanged and the 
supports are immovable. 


17 
















To simplify the proof of this law let us deal v. ith two points m and n 
of any framed structure, and let Sn\v\~ displacement of the point m when only one 
load P„=l is acting at the point n , producing stresses , 

Also let ^nm - displacement of the point n when only one load I is acting 

at the point m , producing stresses ^ 


Then applying Mohr'a work equation (l^D) to the two cases of loading. 


we have : — 


Sj 


Deflection at m = point n. 

Deflection at n = for I at point m . 


3ince S =s :E S 


^mn “ 


Sn.e 

EF 


therefore 


,(15A) 


This is Maxwell's law and may be stated thus i- 
The displacement of a point m in a direction mm' , produced by a unit 

load acting at n in another direction hn', is equal to the displacement 
of a point n in the direction , produced by a unit load acting at m in 
the direction mm' . 

Art. 16. Menabrea'e Law. 1858. or theorem of least work. 


For any statically indeterminate frame, the redundant conditions 
reduce the actual work of deformation to a minimum. 


Now the actual work of deformation for any frame including redundant 
members ie given by Clapeyron's lav/) Equation (12A) as «- 


A = i 

^ eF 


.(16 a) 


and this must become a minimum when the stresses S have the values given by 
Scuation(llA) as t - 


S = - S^X^- SiX^- ,etc.. 

Differentiating Equation (16A) gives t- 

HA SBS. 


.(16b) 

.(16C) 


Also the partial differentiation of S with respect to each X in Equation 
(16 b) gives t- 

•H;" “ ' Hfc* ~ ’ He" - 5 >.^ 

Dividing both sides of Equation (16C) by and substituting 3S __ ^ gives i-- 

, .?.(16s) 




E’F 


rr 


which must be equal to zero if there are no reaction displacements,since by 
Mohr's work e-.uation the displacement <^ss^S , for the point a., which 
is the point of application of . * * FF 

Similarly then. 


^ o •, "M ^ o i M = o 


c+c 


which proves that the redundants reduce the actual v/ork of deforTTlfl^•■^on to a minimum. 


18 



















Art 17 .Castigliano*a law (1879) » or derivative of the work equation. It deals with 
the displacement of the point of application of a force. 

The actual work of deformation due to a load acting at m 

on any frame, is by Clapeyron'a law 

. .(17A) 

where 0 ^ ia the diaplacement within tne range of proportionality, of the point m 
in the direction of 


Also, the actual internal work of deformation for the entire frame 
ia by the same law 

A; = is .(17B) 

partial differential derivative of A with respect to any external 
load is from (17B) 

"3^ .L g S 5 I ^ \ f 17 c) 

wherein is the derivative of the stress S in any member of a determinate 

or indeterminate frame, and may be evaluated from the general equation (llA). 

Noting that for a single load , then equation (llA) gives the 

total stress S for and redundants X as ; 

^ >6#'^.(17D) 

The stress S,„ in any member S for I , is independent of 

the X* s, and the X* s are independent of the load P,^ , hence S may be partially 

differentiated with respect to ^ or any X^and equation (17D) when so treated 

gives ^ 

ana , ate,..(17E) 

Substituting for ^VaP„ in equation (I 7 C) then 

|A = ss„s(^) .(17F) 

But Mohr's law, equation (l^D) gives a value for deflection as follov.’s; 

|.S,^=SSn,S^ . (17G) 

Therefore, 

••S„= (17H) 

which is Castigliano's law and means that " the path Sm of a load is enual to 

the partial differential derivative of the actual work of the frame with rearect 

to the load Bn 1 

It is thus soon that Castigliano's iaw represents an equivalent for 
Mohr '3 work equation except that Castigliano arrives at his result by a more 
circuitous process. 

For solid web structures involving direct stress and bending, Castigliano'a 
equation takes the form. 

..^ 

where N and M are effects produced by any actual c ase of loading 


19 


19 



















Mohr's law, leading to a more direct solution of deflection problems, 
will generally be employed in preference to Castigliano, though the latter did 
receive earlier attention of American writers due to English translations of 
the Italian text. 

Chap. IV . Statically Indeterminate Structures . 

Art If: Indeterminate Frames by Mohr's tVork £~iuation . 

It is now proposed to show the manner in which Mohr's work equation 
may be employed to find the stresses and reactions in any structure involving 
redundancy. 

The application -ill be made first to frames, loaded v.ith auny system 
of loads P, concentrated at the several pin points of the frame. 

In Art 11 it was shown that any problem involving redundancy could be 
reduced to a principal system which must be statically determinate, and then assign 
the redundants to the principal system es conventional loadings along with the 
external forces and reactions. 

For each redundant so disrosed or assigned, Mohr's v/ork equation will 
give a means of finding the displncer.ent of its point of application. Hence 
according to definitions given In Art. 11, for redundants 

etc.,and displacements in their points of application ^ 4 ^, f , -- ..etc,, 

respectively, Mohr's work equation,(14 d) can be applied once for each redundant. 
Introducing the value VeF each member of the frame, these equations are 

£S.|| =£S,S? ) 

Sl,= £St|£ S^S? ). (ISA) 

eta,for any number of redundants, neglecting temperature effects and abutment 
displacements. 

In these equations the value of S is represented by equation (llA), 


for any ease of redundancy, as 

S= S, - - S|,X|, — . (133) 

and this value introduced into equations (18 a) gives as many elasticity equations 
as there are redundants as follows:- 

c ‘ ’ 

2 s. Ss = £ S. S, S - X.S S. 5 - 5. S,S.* ,.»c.) 

Sb=2SbS?=£SiS.5-X.£5bS.^-Xi£S^^-X,SStS.5,el>:.) . (18C) 

SgS ^ Sg 5^ ^ — X«t^ Sg ^ , etc.) 


wherein the summations include only the members of the principal system. 


20 









Now ,etc, bein-: changes in lengths of redundant membere of 
some kind, these may be evaluated in terms of their lengths, areas and stresses 
( or reactions ) and become t- 


* tt: 









etc., 


,(18D) 


For external redundancy with immovable supports. 


S = = <5 a O 


Hence, all the terms in equations (18C) are now knov.n, except the three 
redundant forces ^<».>^-^>and X^, » having three elasticity equations 
involving only three unknowns, the latter may be found by solving equations (18C) 
for simultaneous values of the X's , vvith the aid of equations ( 18 d) « 

.ihen the X’s are found for a certain case of loading, then e.-.uation 
(18 b) v/ill give the stress in any member. 

To include temperature effects or abutment displacements it is best to 
evaluate these separately with the use of the general work equation (l4c). 

Art. 19. Indeterminate Solid V>eb Structures by Mohr* s i.ork Sguation, 


The procedure is similar to the one just given for frames except tV it 
Mohr’s work equation ( 14 e) is employed together v/ith the general moment equat-on 
(11 a). Since the effect due to axial or direct stress is usually neglible, v/e 
will considei only moments for the present. 


As for frames, so also for solid v/eb structures involving redundancy, 
we must first reduce the problem to a principal system and assign the redundants 
as external forces, moments or reactions, and conduct the analysis for the 
principal system selected, iVe will proceed on the basis of three redundants 

^ CL 1 ^ I, > * 

The work e^mations for the displacements of the points of application 
of the redundants are then according to equation (14 e) 


u 


ui 


^ £lJ 

flA 

) 

) 

) . 

) 


'^“eT J 

f M 

.(19A) 

1 ^ » 

^ fl. 

/'MM.dx 

) 

) 



wherein the moment M isilie moment at any point of the principal system due to lof’r^ 
P and redundants X acting together, while the moments M-^ and are those 

moments in the principal system,respectively for the conventional loadings 
X^= • > > Ckfxd X^ = J . 


Now the general moment equation (llA),including the effects of the 
redxmdants, is t- 


M = M, - M.X,- M^Xj- X £ 


,(19B) 


wherein M*ssthe moment at any point of the principal system due to the actual 
loads P, 

Substituting this value of M into equations (19A) the final elasticity 
equations are obtained as follows »- 


- X./M. M. dx ) .(190 


21 











In these equations everything can be determined by pure statics 
except the redundants X v;hich must be found by solving the three equations ^(9cj 

The displacements and become zero v;hen they refer to the 

displacements offix 6 d supports and the redundants are external. If the supports 
are subject to displacements, then appropriate values must be fotind or assigned 
from known conditions of the problem in hand. 

The evaluation of the integrals is accomplished in a very simple 
manner by substitution formulas which will be explained later in connection with 
practical problems. See Art". 26 . 

After the redundants are found from equations (l 9 C),the moment at any 
point of the structure may be found from equation (I 9 B), or any reaction may be 
evaluated from :- 

R Rtf'* .( I9B) 

In the above case three redundants were assumed to exist, but the 
number of equations and the terms in each may be extended by similarity to any 
number of redundants. Also the above expressions can be reduced at will to 
cover one or two redundants simply by omitting the terms which are not needed. 

Art.2Q . Indeterminate structures by Maxwell’s law . 

The application of Maxwell's law of reciprocal displacements to any 
structure involving redimdancy will nov/ be given. This necessitates a 
transformation of equations (18C) in which the summations are nov/jexpressed in 
terms of elastic deformations of the structure instead of tho stresses in the 
members. 

In equations (18C) each term consists of the sunimation cC the products 
of tv/o stresses, which according to Mohr's work equation v/ill represent some 
displacement value S . It is, therefore, desirable to adopt some standard 
nomenclature v/hich v/ill always be understandable. This will consist of double 
subscript-bearing S's , in which the first letter will always refer to the 
point at which the displacement occurs, and the second letter will alv/ays refer 
to the cause,or conventional loading producing the displacement. 

For example let 1 - 

^ma.= displacement of the point of application m of any load P„, , 
in the direction of this load, when the principal system is loaded 
only with the conventional load 

5^^«the displacement of the same point »»» for the conventional loadingX|S|. 

^^-■sthe same for a cause or condition (. 

K. a the displacement of the point of application ft of tho rodimdant 
X,. in the direction of X,^,when the principal system is loaded 
only with 1 . 

a similar 

5 3 a similar 


displacement of the point ^ duo to the conventional 

loading X-^= I. 

displacement of point a for X^a I, etc, for any other 

cases. 


22 





Mohr's v/ork equation (l^D) then gives the follov/ing values 

4c = ^«b > 

^ S * 4t ) ^ ~ ^ ^ ^l> ^ ^C S ' ^hc 


etc.^for other similar values. Hence the elasticity equations (18C) v/ill 
assume the follov/ing form :- 


^a4ui“’'^b4»b ^ac j 

^b = ^ Pm ^mb“ ^l»«.~"^b4l>"^C ^>C ^ .(20 A) 

5^ = ^ x»4a -^b4z> -\^CC. ) 

) 


4c* 4 a 


It _3hou]^d be noted that according to Maxwell’s law # 4b“ 


5,.-4. 


s =s 

ma am 


ehc. 


ba > 


Hence all the deflections in the first equation (20A) can bo obtained 
from a deflection diagram dravm for the principal system with a single loadX^=l. 
Those of the second equation for a similar loading of Xt,= I , and those of the 
third equation for X^= I. Again S^:z 4=0 for rigid supports. 


Therefore, any method of obtaining these deflections, preferably by 
iVilliot-Mohr displacement diagrams or by area moment4 T/ill make this method 
applicable for the solution of the redundants X, After these are found, then 
equation (llA) v/ill furnish the means for finding stresses, reactions or moments 
in any structure with redundancy. 


It should also be noted that equations (20A) apply to any structure 
v/hether framed or solid v/eb. 


Chapter V. Deflections . 

Art. 2 I. The variety of deflection problems encountered in bridge and structural 
designs is so great that we can scarcely do more than describe the methods 
employed in the more involved problems, and then concentrate on the class of 
problems .. ith which we are more particularly concerned in the field of buildings, 

..hen dealing with larger bridge problems and complicated loadings,the 
simplest and most complete solution for displacements of all pin points is 
obtained by the graphic method known as a V.illiot-Mohr diagram first presented 
by the author in a paper road before the Detroit Engineering Society in 189^, 
and published in the Jour, Assn, Eng, Soc, 

Another method possessing greater accuracy and especially useful for 
stress computations in large framed bridges involving redundancy, is the semi- 
graphic method of elastic weights contributed by Professor Mohr in 1875. 

All these have received exhaustive treatment in the author's treatise 
" The Kinetic Theory of Engineering Structures ", and will not receive further 
attention here. 

The method of area moments affording either a graphic or analytic 
solution, was first developed by Professor Mohr in 1868. This is a very useful 
general method for solid web structure^ and should be thoroughly understood. 


23 






The method is especially useful when the entire elastic curve is desired for 
complex loading and variable moment of inertia. 

The most comprehensive general solution of deflection problems 
pertaining to solid web structures is by Mohr's work equation (14E) and eq. (14 f), 
The same law expressed by equation ( 14 d) for frames^gives’^^^deflection of a single 
point only. 

Art. 22. Deflections of Beams by Area Moments . ( Mohr. 1868. ) 

The deflection problem which it is proposed to solve may be stated as 

follows j- 

For any straight beam simply supported on two supports and loaded in 
any manner by loads normal to the beam axis, to find the deflection at anj' 
point m of the beam, in a direction normal to the beam axis. 


The solution is accomplished by drav/ing a moment diagram for the given 
case of loading, then treating this moment area as a load area ( moment ordinates 
in kip-feet x lengths in feet ) and finding the moment of the moment area load 
about the point m where the deflection is to be found. This moment divided by 
E I will be the deflection provided E be taken in kips per square foot, 
and I in ftf in conformity with the units chosen for moments and lengths. 


Problem 22k . 

The solution of a problem will best illustrate the method and we will 
take the simple beam with a uniform load p per foot to find at the center , 

The moment area for the 



uniform load p 
with the middle 
m aaual to 


is a parabola 
ordinate aw 


The area of this parabola is 
equal to 2/5 times the length 
t in feet, times the middle 
ordinate p£® in kip-feet. 

^ ,(!> 




2p' 

Z4 


Hence the end reactions A and B for the moment area will be equal to 
and the resultant of the half area of equal weight will act 

through the center of gravity of the half area at 5 £ to the right or left 
of tht3 center m. 1^ 


1 ^* 


Renee f- 



Should any portion of tho moment area be negative, then the moment, of 
the negative moment area must be negative with respect to tho moment of tho 
positive area. 


24 






















Problem 22 B « A simple beam with concentrated load to find 
under the load. 



the deflection 






Pal> ^ zb 
TT" "3 


]-r[# 


t- 


Pa’l 

TT 


”+* f^jzh+Sabi-aJ 


Hence S = ^ 

"* 3FI £ 


The solution of these problems is Ion thier when dealinr, v.ith lettered 
diminsions than with actual numerical cases. 

Proof of the above analyt,lc solution follov/s very simply from an application of 
Mohr's v/ork equation ( IhE) which is 

M^dx. 

In this e ;uation Ma is the moment at any point of a beam due to a 
unit load at m , but it js also equal to the moment at w due to a unit load 
at an;y point of the beam. Therefore, represents the moment at m due 

to loads M clx on any differential length dx and the expression M must 

represent the total moment at »ti . Hence this total moment divided by E I 
must equal l*S„, according to Mohr's work equation (l^E) thus establishing the 
basis for the method of area moments, 

Ari,3 3 . Graphic solution of deflections by Area Moments , Mohr, 

The problem is exactly the same as stated in Art, 22, and the solution 
is now accomplished by applying the well known graphic method of finding the 
moment of any set of forces u) by means of a force polygon,and an equilibrium 
polygon. The forces in the present graphic solution will be the partial areas 
of a moment area and may be called elastic weights tu s Mdx for M in kipft.^aud dx 
in fact. 


The properties of an equilibrium polygon are such that the moment at 
any point m of a beam^is the ordinate of the polygon at m,multiplied by the pole 
distance H employed in the force polygon. 

But since the deflection is moment of the elastic weights ia; 

divided by El, therefore, if the force polygon be constructed for weights iat 
with a pole H ^ E I, the ordinates to the equilibrium polygon would represent 
the actual deflections to the scale of lengths chosen in drawing the beam. 


25 



























Now since these deflections would be too small for proper scaling. 

It is v/ell to chose a pole H a El ^ for which the deflection ordinates 
would be 100 times actual 100 
to the scale of lengths. 

Problem 2^ A . A simple beam on tv/o supports, loaded with loads P, to find 

the deflection at any point m of the beam. 



The 1 moment diagram ( 9L ) drawn for the loads P, with pole 10 kips, 
gave moments Vio times actual to the scale of lengths, so that the actual values 
may be written on the moment area by direct scaling. 

The M area is nor; suitably divided into partial lengths dx and the 
computed values of the w leads are then applied at the respective centers of 
gravity of the partial M areas. 


’’he force and equilibrium polygons ( "b ) are then drawn as Indicated 
in Fig. 25 A, using a pole H s E I thus obtaining a deflection polygon 
with ordinates drri which 100 

are 100 times actual to the scale of lengths employed for the span £. The 
entire elastic curve is thus found for the case of loads P as shown. The actual 
value of under Pj is found fe be MooX7.5ft. sO-crsff.s 0.90 inch. 

In cases where the moment of inertia is variable, it is best to choose 
the elastic weights u)s Mdx and then employ a pole H r E to obtain 

^ ’ 100 

dofloctions lOO times actual to the scale of lengths. 


No further proof of the graphic method is deemed 
principles involved are identical with those domonstrated 


necessary, since the 
for the analytic method 
in Art, 22, 


26 























































AfjK24. Slope of the elastic curve by Area Moments 


Since the slope of the M moment curvo^or polygon^at any point of a 
beam is represented by the moment dorivative,+hkf»foiV d M ^ 


■37 




But dM/4jjS3 Q 7s the shear at the point where the moment is taicen. 
Also since the deflection Bm is the moment of the elastic weights iv about the 
point m , divided by E I, therefore, the shear of the elastic weights on one 
side of the point m , divided by E I, must equal the tangent of the slope 
angle to the elastic curve at the point m , 


Also, the tangent of the elastic curve at any support must enual the 
end shear or reaction of the elastic weights ur divided by El. 

In Fig.25A, the end reactions and R-^ of the elastic weights are 
found from the force polygon of those weights by drav/ing a line from the pole O , 
parallel to the closing line A'S**, thus dividing the total weights iw into two 
portions representing the end reactions of the weights iv . I'dth known, 
the shear at any point is easily found,giving t- 

, .( 24 A ) 

This applies to either the graphic or analytic methods. 

ATi.2S.Deflections of frames by Mohr’s work equation . 

Mohr's work e^-uation (l4 D) affords a raeans of solving any of the 
following four types of problems for any statically determinate frame and any 
case of externally applied loads, producing stresses S in the members of the 
given frame. 

A. To find the displacement of anj^ point m of the frame in 

arx''^ direction, by assigning a conventional load • in the particular 
direction in which the displacement is wanted. This represents the conventional 
loading of a point m 

3. To find the relative displacement between any pair of points m end 
of th^ frame, by assigning, a pair of conventional lotds 1^* • > acting in 
the direction m-m, . This represents the loading of a pair of points m and m, 

C. To find the angular rotation of a line m-m, joining any two points 
and m, of the frame, by assigning a conventional unit moment to the linem-m^» 

iVe call this the conventional loading of a line m-m,. 

D, To find the change in the angle © between a pair of lines fn»ni,and 
n-n, of the frame, by assigning a pair of convantionsl unit moments to the frame, 
one for each of the pair of lin;s. This represents the conventional loading of 

a pair of lines. 


The four cases will be illustrated presently and each convention will 
be referred to as a loading ^ the of which will be the path or elastic 
displacement of the conventional loading. 

If the supports undergo known displacements at as a result of the 
actual loads P , this affect must be considered in evaluating . 

Likewise temperature displacements may be found by applying temperature corrections 
to the aC’s of the members according to equation ( 8A) . For the present, 
however, these effects will bo neglected. 


27 









In the solution of problems it Is very important to observe the signs 
of all stresses eind A Cs . To avoid confusion, the positive sign v/ill always 
be used to denote tension and elongation, while the negative sign v.ill denote 
compression and contraction. 


Problem 25 A » Pisplacement of a point w 
I -0.300 3 -0>450 5 


of ,any fr^e . 

6o\: 



stresses in kips. 

A 


Given a 
displacement S„, 


lo 

simple truss on determinate supports. Fig. 25 A, find the 
of the point m , in the direction of the unit load 




when the frame is supporting some actual system of loads, producing stresses S ^ 
and changes A^= Sf/fF in the lengths of the members. 


The solution is effected by evaluating Uohr's work equation ( l4 D ) 
for the particular example given in Fig. 25 A, using the conventional loading | 
of a point m . The eauation is t- 



Compute the value a£ for each member, for the total stress S , length 
t and sectional area F of such member. Also, find the stress in each 

member by drawing a Maxwell stress diagram for the conventional loading | . 
Tabulate these values a. C and as in Table 25 A, carefully observing the 

signs, + for tension and elongation, and then compute the products C , 

the sum of which covering all the members, will give the desired deflection 
in the same length units as used for the a£'s 


Mem 

. 

s 

F 


a£ 

^ w ■ » w 

Sa 

Sa 

a£ 

Mem 

S 

F 

e 


Sa 




K’fPf 

$4-in 

in. 

in 

Kips 


— 


Kips 

Sq-in- 

in 

irt. 

Kids. 

4 

— 

0-1 

-245 

33.5 

4-62 

-0.1126 

-0-235 

0.0265 

- 

1-2 

+ 68 

13.7 

360 

+0.059S 

0.0 


- 

1-3 

-2S2 

26.5 

290 

-0.0920 

-0-300 

0-0276 

- 

9-10 

+68 

13.7 

360 

+0.0S9S 

oo 

? — 

- 

3-7 

-284 

29-4 

560 

-0.1864 

-0.45O 

0.0640 

- 

3-4 

-44 

J4.7 

360 

-0.0361 

-0.188 

0.0068 

- 

7-9 

-2S2 

26.S 

290 

-0.0920 

-I.09S 

0.1008 

— 

7-8 

-44 

)4.7 

360 

-0.0361 

-0.690 

0.0249 

- 

9-11 

-245 

33.S 

462 

-0.1126 

-0-870 

0.0980 

- 

5-6 

- 68 

14.7 

360 

-0.0055 

0.0 

— 

- 

0-4 

■I-IS8 

15.9 

580 

+0.1925 

+0&A0 

0.1230 

- 

1-4 

+156 

17.6 

46Z 

+0.1368 

+0.235 

0.0322 

- 

4-6 

+2S2 

26.9 

580 

+0.0906 

+0.600 

O.07ZS 

- 

8-9 

+156 

17.6 

462 

•K>.I368 

+0.880 

0.1202 

- 

6-8 

+252 

269 

580 

-1-00906 

-►1.095 

0.0993 

- 

3-6 

+ 48 

17.6 

462 

<0.0424 

-•0.235 

0.0100 


8-11 

+158 

15.9 

580 

Id). 1925 

+0.550 

0.1060 

- 

6-7 

+48 

17.6 

462 

•^>0424 

-0.235 

- 

0.0100 

Forword 

0.7377 

o ■ 

L 

0.9318 

0.0100 


in ike example all siresses are in kips, E s 3o,aoo kips per leiK^iksare in inches, 

and Cross-sections r m s^. inches. 

Hence 34-0-93 I8-0.0|00 s-h0.9218 inch. 

ffl 


28 
















































Problem 25 B . Relative displacement between any two points m and tY\, of any frame , 

Ti) O (3> +0^32 (5) +0.52 (7)4-0.63 { 9 )^ ! 

-Px I kip 

S_ stresses *m kips. 



F»§.25B. 

The example in Fig. 25 A is used again to illustrate tae solution of 
this problem where is nov/ the change in length of the distance m-m, 

which v/as produced by the changes in the lengths of the members. 

See above Table 25 A. 


The conventional loading is that of a pair of points m nd m, and consists 
of two equal and opposite loads Ps l applied at the tv/o points and acting in 
the line m-m,. In Fig,25 B, the unit loads-are taken as acting av.'ay from each 
other on the presumption that + would bo an elongation. Should the value 

of turn out as negative, then our assumption was v/rong, and the arrows should 

point toward each other. This does not affect the solution of the problem, but 
merely corrects’an erroneous assumption if such was made. 


The stresses S«- for the two unit loads are again found from a Maxwell 
stress diagram, see Fig, 25 B, and the computations for are then carried out 

as in the previous Problem 25 A. 

The answer is t- 

S^s-hO. + 0.0 3 incK. 


Problem 25 C . Angular rotation of a line m-m, , joiniag any two points m and iTij 


of any frame . 


'5»o.oiz 8 kip. 
.m. 



m-m. 


stresses in kips. 

-0.0O53\ ll 

Rs+o.oo 69 Fi^.25 C' |R=-0.0069 kip. 

Again employing the example in Fig. 25 A, the rotation of the line 

will now be found. 


Qe will assume the same changes a £ in the lengths of the members and ascertain 

the rotation 5^, of the line joining the two points m r.nd m, . 


The conventional loading is that of a line m-m, and consists of a unit 
■oment produced by two forces •/« acting in opposite directions through n lever 
arm e , and the rotation is assumed to be counterclockwise. Should the computed 
Talue a£ come out minus, then we would know that the rotation was actually 

clockwise instead of in the direction above asaxmied. The value of 5,^ is expressed 
aa the tangent of the angle of rotation. 


In the above example, the S* stresses as written on the members were 

obtained from a Maxwell diagram dravm for a moment ® — I -, and these 

atreasea multiplied by the of Table 25 A give the products S^a C,^ twelve 

times actual because the unit moment should have been 1 kip-inch i instedd of 1 kip ft. 
TTio answer thus obtained is:- 

I Z + o. 007«Z - O. O OZ58 =H-O.OOS24 ^ or 0.0004-4, 

which is the tangent of the angle of fotatlon of the line mm, • 


29 





























Problem 25 D . Change In angle 0 between a pair of lines AC and BC , 

of any frame. 


The solution of this problem is illustrated on a three-hinged, framed 
arch, and involves the conventional loading of a pair of lines AC and BC^ Fi^.25D. 



The stresses 5 producing changes ^ C in the lengths of the members, 
which in turn cause the angular change in the angle Q , are found for any 

given case of loading and tabulated as in Table 25 A. 

The conventional loading consists of e unit moment applied to each of 
the two lines, and the directions of rotation are so chosen that the angular 

change in the angle 0 , is a positive increase expressed as the tangent 

of a small angle. The stresses , due to the conventional loading indicated 

in Fig, 25 D, are no7/ found by drawing a stress diagram or by computation, and 
these stresses added to the tabulated values for all the members of the 

structure, furnish the means for computing 

The numerical solution being similar in every i.ay to the previous 
examples, is not carried out in this case. 

Art:.26.Deflection of solid web structures by Mohr's work equation, applicable 

to any principal system. 

'.'.hen dealing with solid v/eb structures, the work equation takes a form 

which requires the integration of the products of two moments, over a distance 

for which the moment of inertia of the cross-section is constant. 


wherein :- 


work ecuation (14 d) is 
‘ft 

/ M 


m 


Ms the moment of the externally applied loads P 
of a principal system. 


about any point m 


. the moment about any point tr\ due to the conventional loading 

the deflection at the point m due to the applied loads P 

If it wore not for the integrations involved in the solution of problems 
by means of the above equation, this would afford the simplest and most comprehensive 
general solution of deflection problems ever proposed. 


Hence, the removal of the only deterrent factor, viz j- that of integration, 
will enhance the usefulness of Mohr's work e nation and make it possible fo" anyone 
not familiar with the calculus, to solve quite easily any of the ordinary deflection 
problems and redundancy problems pertaining to solid web structures or rigid frames. 


JO 












TABLE 26-A. EVALUATtOn OF THE EXPRESSIONS ond/^fA^dK. 

__ Moment of Inertia constan-t over Icng-th £. _®_ 


rpi-, / MjMydx = fry. 

- ! /Vdr = fzV 



-‘dx 


MyOrea y 


area y /t 2 ^ ^ 

^-'i M^dr=fy. 



Mj^area 


1 

1 

1 


1 

4 ,o.syp-«^ 

iA t 

L- P 

1 

— ^ 


My dx s 


/■'M^dx = |y_^ 
areas ^ yj 


'4'["'-I'] 



y VWydx=|z,y, . 
7Vdx=lr* 


* ®/^ Z Q 

. / Mydx = l.y^ 



M Y dx=i jy fai-b') +.5.^1 
^ ^ J z 2^2 

LZ ! ^; 7''M>4y^ 

H- 1 -y y s-'i 

•yo 



y M,M,dx-t[ry,.yy 


X My area ^1 


M. 


'^’‘=l[y>xyxV/] 



^ /M.M,dx=&(|,,.z,) 

7 Hdx=^ 

-<o 


My area = ^ly . 


M^area 


1 / M.M,dxx|[r{ 2 y*^')*;^(y«y^| 
3 . ? r X . . ., -, 


1 I M7x=i[r7z !,*<]■ 

I ^ X. 


X My area ^ . 

L-e —i 7 ^‘''‘4[y>>;yz-yr ] 



yM^M,dx = |fr,y, 

M;dx=8fyX 

y = A(rfx-K^). 


F/ M^oreo 



r yM,Mydx=:|jf(r+z^).^bz^a^J 

when z=o, yM^Mydxs 5iy(e+a) 
j when Zj^sOjyMxMydxa^iy^^-tb) 

■b — y Mj,dK = .|.7 





2 1 


,M^dx = |er,y, 
Mjdx=Aeyf. 

yx2^(e-x). 


To approximate a curve 
by three ordinates. 


grecT"""^ f 



Z 2 


M_arca ^2 


My dx= A ?y*: 


-1 area -^ty. 




M.M^dx-H[f+^]. 
Mjdx = .a£y*. 


' M^area 


14 

-^ 2 

*'MV1y0rea 

» (renerol case 


^/z—n- ^2—n 


y^M^My dx s^ r, m ,J 

3 Sta+ic mom.of MyareaoboutH 
' m.sStdfiemorn.ofMvareaaboutZ '2 
' ^ ' 
i j MydxssZm .where 

J-A m = static mom.of My area 
^ about A-Aaxls. 


Note: Incase the My and moment diagrams have different lengths f,then the integral 

Can Cover only that length which is common to both areas, because the moment ordinates of the 

shorter area are sero over that portion of the length which is not common to both moment areas. 

The ordinates y and Z are all positive in the above formulas, and whenever negative areas 
or ordinates occur, they must be introduced with negative signs. For o negative area.,the 
integrol Jt/^d% is a positive <yuanttty. David^MoUtor.C.E. 































































































+ha+ 

After some study the 7;riter discovered^these integrations could be 
performed once for all, thus exhausting all possible combinations of norier.t 
diagrams v.hich might be encountered. 

I 

f The Table 26 A gives the evaluation of the integrals^ 
and M^dx , for all possible moment diagrams and .vill prove quite valuable 
as a practical expedient in solving numerical problems. 


The formula^ may be called substitution form u las , and their use .ill 
be illustrated in Chapter- ^ by solving a number of problems. 

CHAPTER 6. PROBLEMS BY MOHR'S WORK gOU/^TION . 

Art.27» Dgfleciion Problems. 

Problem 27A. Tindthc deflection at any point m of n simple bcam^^ 

)f a uniform load p per foot of Sparx, u4in<^ fte formula l•^„=^^ . 


for the case oi 



Sol tiflon . Sketch fhe M area which is 

a parabola. wiiU midtile Ordinate « . 

Skcick ihe area , which ts a irtant\le 

With max. ordinate r |. , rio.27>^, 

Vow employ the substitufion Formula 

Table 26A , Case 7j rvakino 

, .8 

Z= . Th^n we ha\/^c direcfly 

ZAEI ^ e J 24EI '' ^ 

This I^lvcs at once the elosfic curve for ihe def {ertlon at any paint nn. 

The maximum deflection at the center occurs when L tYxakinq S = 

2 . * “ m 3g4£j 

Compare this with Problem 22A where the Some Case is solved by area-moments. 
Problem 27 B. Find the deflection at m under the concentrated load 


acting oh a simple beam AB- 

K - - a-jSl’--b 



since 'the conventional load P= | now 

aefs at the load point m, Fa- 14 F ii applicable 

a I Vina S = — fwi^dif.- 
V T m p^Eiy 

Fi ^.276 shows the moment clia^jram wifh 


the moment ordinate underthe load point m- Hence from Table ZSA, Case 5, 

JVlydx = ^y ^ivimj . Compare this with Problem 22B, 

Sol^^ea by area^-motnenfs. 


32 































B 

v i r* * 




-P-3'- 





ProbI em 27 c . Find fhg horizon^'al an^ 1He t^grtical dgftecfions ofihe poinfC for 
the column wifh cantilever arm BC supporting a verfical toad Pat C . 

Q. Solu-t-ion for fl>ie vertiCAi def 1&ctlon of Ihc pomt-C . 

Since ihe point-C is the loaded point^andthc required detl. 
Is in the direci’ion o-f P, mai^apply the conY^r^t■iono,^ lood P 
and use E<j. I4F Which is ‘ 

The i-rtomcnt diac|rom tor P is drown in Fi<).Z7C,ilnd 
+he inteqrol JwC’dn iseyoluat’ed for the height h and the 
r«ii<|th 2 of-the caniilere.r arm- 

From Table 26 /t, case I, we obfo m ior ihc colum n •- 
J^uC'da^ 'hz*‘= lo(3P)*k 90 P^. 

Also from Case 2 o-f- i-able^for ihe. cantilever arm 

/Vd«=iz*=|.(SPf=9P‘ 


A 


-3P 
•- z -> 


Tr^TTmrrrfrrrwirrr 


Given Cose M cliai^ram. 

Fig.27C. 


For ps4 kips , E aSo.oooz I44 =4, 3ZO, OOO kips per and « 1^^= 12 mokin^ 

EX*2Soo in kip foot units, then we ' hav-e for-the vertical defleefion of C , 

K = o. isa4- ft. = 1-90 in. 

* ZSOO 

This deflection nc^jlects the shortening in the column due to the direct stress P, 


Ta. Solution for the horizontal deflection otthe point C . 

We must now apply a convent ionol load P» I , horizontally at C ^ and em ploy F^. I4d, 


which is y'M M^dz . 



The moment diagram ior? is the same as 

before, tut the dia^rom -for a horizcrttaI 

load P=| at C, isatriani^lc with base h^lOft. 

Since the area, is zero on the Cantilever, 

the inteqrationyMM^djt applies to 4he column only. 

From Table 26A, Case “i , we obtom the value 

yMM,^dx. =:!^(zy^ + y^)= »Sii£{6P+3P)^ isop, 

makinq S. » . which for P = 4 kips and 

^ ^ FI, 

FXsZSoo os above , then 

S = .±5i£5_ s. o.Z4ft. = 2.88 in. 
h 2900 


33 



































Art. 28.The Th re e-Moment tCojuation and its>Appllca{rloni> . 

Problg.m 28A. Proof of the Three-Momenf Ct^uofion ,Uniform Loods , Vonablg I , 


par ft. y t?, perft. 



-ct;-: 



R 

• t ^ 


R. 


Given 

r * 

case. 



- ^17 

IL_ C, - 




__ c,- 


Principal System -2 simple beams. 

Mj and MjUre regarded as known tfKtcrnol moments and n a reciondont to be eraluoted. 


R, 


(a.) 


-M: - 


-M, 







-4!-t, — ^ 

CondttianX®o M orcas. 

^ O 

The M^area Consists of 4- separate portions^ 
2 -ne<^at« re areas and 2 -posifive areas. 


I, 


X?' 1 


Ix 


r 

it 


U- - C. -- 


- - L - 


Conclit:onX=-.' M areas. 

a ’ a 

The X^-l momen+ »s produced by a unitlaod 
octing I ft. from the enel of aoch Sport 

(d; 


The etosficify Eguafia n fer one redundant condi'fion , occordm^to E<^.I9C is; 

I.S^si^ =c -for on unyie)din<j support E’is Constant and Cancels. 

Usiriq the formulas Case Z ond Cose S from Table 2€/^, vre can now eraluote the two 
trtr«/|rols jobservln^ t’haty'^M^ fVl^dx must be made to corer the two neifative ond the two positive 
M^oreas, comP.i.od with the area of the same span - This (fives at once. , 

* = + i;» X U -Am - A M, 

ry * 31, « 3i^ fi 6i, ' 6i^ i 


Ond 


ry“*‘^’‘“s*^4' T*) ■ 

These values s>ubstttuted m the above elasticity e<fuatian qiire : 

H •+ — M, ti _ Ji ^ >C« / A' + A ) ^ -_—_ 

MI, Ml. 61, tl. S ll, T. ^ 


(I) 


_( 28 A) 


Multiplyinq E(^.l by 6 and transposin<}, then 

M,i.' r 2X,(i.'+ i» ) + M. !■ - **» ^2 _ 

'I, * 4 . I.' »i. ^T, 

whtch is the Th ree-Moment equation for Uniform Loads , The positive si^ns merely prove 

thot was properly assigned in Fi/f.(h). 

KnewirKf that the three ferms of the ie-ft hand side of Eq.ZSA represent negative 

moments, therefore , if this eifuation is to qive the proper si^n «v« must place the ne^jative 

siqn before the fwo ri<fhf hand fermS and obfain the final equation os . 

Z M (A^. Aw M- A = - AL^- AA -(28B) 

*VI,. ij 41 , 41 ^ ' 


M. A 
‘ I. 


34 






































Problem Z6B. Proof of the Three-Moment Equofion, Concenfrafed Loads.VQrtflble I . 






M, 


k,e,' 




R. 


e.- 








D C 




/C 


R 3 


Z.- 


P^ M, 

KZ% 




Given cose 


Principal System- 2 simple beams. 
M.and M'x are regarded as known ex.ternal moments and M s is a redundant be evaluated. 

(a) (h) 




Condition X«0, areas. 

The M^arca consists oi A iepara^e portions, 
2 >ne 4 ati«^e areas and z-positive areas. 

(f') 


Condition X=-I, areas. 

The X^= I tnomeni is produced by m unit load 
actinq I ft. from the end of each span... 

(d) 


The elasticity Equation for one redundant condition , according to E<^. I9C iS; 

I.Ss-L/’M M d)i - .2i± / M^dx = O for an unyteldfnq Support R. £ is constant and cancels, 
a Ely ® ^ T • 2 

Usin^the Formulas Cose 2 and Case S, from Table 26A, we can now evaluate the two 
intc<jrols, observing that M^dx must bemad* to cover the two negative and the two positive 

K areas, combined with the area ot the same span . This <jives at once , 

^ ?. KO-^ P. 

and ^ ‘ 

These Values Substituted m the above elasticity e4juatIon,andnotin^ ^l-k)^l-i"k)s: l-k, give: 

— !lili _x.= o-0) 

el/''- “ 61.^ '^ ' sr. el. » '■I. T' 

Multiplying £g.l by 6 and tronsposini) , then 

M,i.' h) = F;k,(i-hf)L + - (^4 

I, *1 *1 

which is the familiar Three - Moment Equation for Concentrated loads.The positive signs merely 
indicate that X^= P>"operly ossigncd in Fig.('bJ. 

Knowing that the three terms of the left hand side of £g. 28C represent negative 
moments therefore, if this eguotion isto give the proper sign we must place the negative 
sign before the two right hand terms and obtain tha final equation as*. 

M,ii+ZM,(il+ i?)-*- M,|.* - («d) 

I, V i, i, J-X 


35 























































THREE MOMENT EQUATION, 


IrCONCENTRATED LOADS, 


V, 

X .X . 

p 

--V; 

. < 

V3 

4 ^ 



I, 

m n 

^2i I:. i : 

3^_ f 



v_-a=k,l,-. 

k- 

-■b,=Ci-k,U,-^ 

r; 

% '• 


General MOMENT EauATioMs . 

M,i' M, Iz = -F^ =-5Plk(l-k‘)l ..-(1). 

-( 2 ) 

K)- M.O-k',)- M.k] =^[f?a>, - M.b.- M^a,] -'-- -( 3 ) 

Posittvc momcnfs produce pension on+he lower side of beam aruicompr; on upper side- 
Negofjvc monaen+s produce-f-ension on fWe upper side of beam and compr.on lower side , 
SHEARS. FirstSpon. 

V = M?~ 4- (l-Vc,^ for poinfs from 1 fo m.-( 4 ) 

V, - t= _ FJ k| for pomfs-from rrxioZ. -(5) 

SHEARS . Sccontd span . 

^2~ for poinfs from 2 fo - (&] 

M, , M^^and are subsfifuted wlfh fheir proper si^ris. 

ReACTIONS . R,=V.+V,' ; R = Vj-hV^ ond R,= V3+Vj .fi+c. - (7; 


HtUniform loads . 


- 1 , 

General MOMENT equations . 

M, 1 4 - 2 M^(l' 4 . li) + Mjlt = - . 1« - \!id ^. li = — s 

M„= + '^(l-x)= M,+V,x- max.when x=^ ---tS) 

1 Z ^ * 

^ I ^ M.^Mz , ustn<) proper signs for M, and ___ __(loj 

SHEARS : V.= ur.L . V^^V-vojl, V^= Ms-Mi , . V=vlu;^l^--(ll) 

y/^ a: Fv1 1 ^ wbicb becomes zero Eor _(l2) 

React l0N*S asEor concentrated loads. ' 

three moment equation for all load effects COMBINED : 

M, i +Mjl^ =_'gPlk(t-k’')l_2m(3k~l)i -(13) 

The m momenf S are positc ve when acting doekwise in f he right hand Span and 
when acting coun+erclockwise inthe lef+haiid span as shown. 

X.MoJiTor ,Apr.i<^ 



36 



















































Probigms28C. Appircafions of the Three-Moment E^uofion,. Consfgwf I . 


Q&nerai E<^^a^^^on>Ce^»^centrated Loade r M,E+ 2Mj(?,-•-2 FkZ\l-^) _0) 

GeneralEgMation-Uniform Loads - M, E,4'2 Mi(E,+^x)+M3£^=-^ -C^) 

Problem A. Beom fixed at owe gnd> simply supported at tiie other. 




r- ^-f-b=KE-n^ ? 

^3--i—-1 R A i 

L p p**" ^....i 

^-- .^s-. ^ J 

f-«-T 1 

1—^ 


In+hc genera/E^».make and M^sO and ►✓ri>e-Ihe new agUa+ion ^or/Vt,^ at A . 
Then from Fig (a), 2 (f+oj s-PkE*^(l-k*’) Also from Fi<)(Zfl)^ 2 oj= — 

<^iVlng M^ = - j=- P|b +bj giving 

Problem B. Bgom on 3 Supports , not restrained at the out-er support i. 

P 


-a=ki, 


f 

0) 


b. 


t. 


-I-*-h 


(^) 


(3) 



(Z) 

Fi^(b) Fi^( 2 b) 

hithe general Equations moke M.sM^ro, and cipply with middle term at (i) as written : 
Then 2M^(f,-g=- fjlt,P.lc.f^(l-kt) Also forffq(zb), ZMjf,+ 
giving M Ptl^t^*{*~l^i) t- Pxkx^x (>•■ hi.) 

Problem C. Beom fixed at both ends. 


giving M — •*- bt.Ei. 

^ ^ ^ BCe.-ej 





t — 


p per ftV 


L-Me 


1b 


Fii(c) 

Apply Eg.I,first with middle term otA for E^=o, 
ond again with middle term at 6 for o, 

This gives the following 2equations: 

2 M, - Pk,£V k:) =- 

M.f*iMj(;«>i-pke‘(i-k*)*- -- 

These are solved as follows -. 

M. + 2.oM,= - P«b(f^0} 

* . E*" 

Subtracti ng •• 

I.SM, =- .^(2E-r20- f- b) 

2 2^*-' 

giving M^=: - ^ (f-b4-2o) s- 
and M, = - Pob /P-a-.2bU- Pob* ’ 

Also M « k^. - 4 ‘l= Pfii/f-o-t*j= 2 j^ 

- T T T TL r f*j t 

37 


Fi^(lC) 

For uniform load M, - So that we apply 

Eg(2) nfiththe middle term atA for tsoandt^^i. 

Then zM.E= -^ * 3M,e,or M=-H' 

Also + 

« 8 12 ^24 


>L»- 


















































Problem Z8D. Aclosed 

by the Three Vlomemt Equa 

^ Ml 

!b per sq.ft 

Ml 

3 

a ? 

1 

VO. 

. - e 1 > 


cf' 

1 

) 

e > 
Pperst. 

‘Me h 

. e - 

^ Ft. 



CJ 

b i-^z ,M,_ , 

i 

Mj 

per sq.ft. 

Mz 


Z8B, 


Given fhe unif loads p, w, and e , on Hie 4 sides of 

fhe-frame,-find >hc moments M, and at the Corners. 

Let ■= P = E and vo 6 _ VV. 

Now apply the S-moment e»^ua-tior\ , first i Hi the middle 
term at iu and oaum ivitht/ic middle terrM at a, und obtoin *• 

M ^ 1- 2 M + M, i - ^ _ £b< -_2VV L’_ 2Eii 

^1 ^1, I' T '^I4T I T ’ 

and 

M. Jl 4- ZM.fJl 1-1 ) +- fll.l =-7.rJ - 

■r • , r ^^ 3 . h ^ 4 I, 

which e<^at"loilS s.m plify intc the f oKoivc’n^ and arc easily sol vcd -forM,and whe n name rical 

volues are inserted. 

Moments of inertia in these situations r^ay, 
be m any units , since eoch term involves Lz8E 

J 


t 4 !; 1 M, + ^ M, =- 2 W f - 2 Eh 


‘ L ' L I. 


h M, 4 (^E )M, = -ZP| - 2Ei [ „ i 

ij i, J-2, ' A, -l-S I 

When r = =■ I3 or I Constant , then the above equations qive 

M - (gh^-r p (^f-*-Zh) — 'h(>iv'^^-«-g h^) |S/^ =—~ -♦■ P 6^ _^gp 

' 4-h*^ - 4(3f-»-2h)^ ^ * 4h 

When p= w , then M,= = ^ILlL £l_ . Haviuq found M,and M ^ ,the 11 the other mom- 

ents are: M^=P?_ly 1 •, m'= M,; M 

ole * _6 ' ’ c g e 8 

We »vi II now Solve the case here shown. 


728 


tr 


Numerical Excmiple * 

IP = 7 3 8O lbs. per 5<y -ft 


'To 

' 1(0 


-M, 


I.--=j_ 4 i^ 


^' 1^3 


^-c t‘?.-o4«- 

(5;940 I '•■ 

>.. 1d5d' 


.V 


I’ 

1 '^ 

l'« 

He 

•Wj 


- r- 


'3-. f® 


. 2 


Im;. 

v> ^ 
lo 


3" 

x^y' 


ijbojn’ ’' Jo'^I> ^ ^5, 

I ' ' 'iL 

-^- 


w = 27-40 lbs c' loc^it. 


e 

—^ si 


-37300 fMbs 


)^= 86300 ff lbs. ^ 2 : 14 900 ft.lbs. 
ft ^ a 

'5L^ = O940oft.lbs. I Constant. 

® j S 3 

P t =ll,C800OO;ch =1,424.ooo-w^=l3_4SO,ooo 

M- >3.>o4.000ic7S’-I^074,000>I2 , 

' 4ki44 - 4 » 75 »75 

M =—Z5. M - l3.304,oaO __ .<s4-QOO-ft 1K< 

^ 4»I2 

M^= g6,3oo-37,30o=+4qoooft.!bs. 

M^= 99,4oo-44-,OOo= 4- 55;4oott.lbs 

16 . 900 - 3.7300 •^44ooo_ _ 5^,750 ff.ibs. 

*= z 

The rcmforcir.q steel for these moments 
is ca[eAAtah?(i tor f < I4 ooo*q" f" GSO^a" 
usinq the formula for ’ 

Inhere M IS m ft lbs, and effect depth 
d = 2 (" for all sides and 29 it Comers. 
The moment Curves can be drawn by ploHinq the 3 moments as shown and the po its of contra- 
are thus located. Nc te that all bent bnrs arc alike, a condi fi on ofte n a ttamable and dear zb Ic. 

^Thoabove cxamplt u a culvert i-.ndcr a railway embank-n’ient with lift of earfn ^ill overthe 
top with Coopers E50 loadmEj on the t rack . The laferni earf-h pressure 5= i»'H ton*'(45-5/j,J Jbi p Sy ft- 
Th iSqiees ft==35H tor ^-3o ,nErioSlbs.p.cuflandH = heiqhtoftillin feet, meJud •••'J surcharge for 
LiveLoaa. Dav^i ci A. Mol I'toi;, 9/226 


P p i 

rt Al . 

N O' 

' I T ,1 
i *► * 

; 


I \ ‘ 

“J4400<^ * 


o -M, 

o ^ 


17 - 0 ' 


38 































ProblgmZSE. Double Culvcrf,Uniform Loading, 

m- ft Fill at 80 Ibs.p.Cu ft. 



numerical Equat'ion4 by substitution of values fromfigure . 


by Three-Momenf Equation • 

Unitstrcsses -fsisooc* f = 6SroV' 

Is * t 

^ Constant, 

Load olatfl- 

psIlZoV, 7700 -ft.lb 5 . 105,000. 

Wslfcoon', 9 i 40 ff.lbs.!S^- 117 , 200 . 

• ' ^i> 

c = |S0 8D, si!— I3J ooft.lbi.ejj -ziA.ooo. 

8 ' 4- 

The probleminvol»'es 6 unknown 
moments ot whichtwo pairs are e<|ual 
by symmetry of loading anddimensions. 
Hence we sol/e tor 4-moments by 
wr»tin<j 4 ecjuations with unknowns 
M,, , M j and M4. This is done by 

applying the 3-inomentequation tour 
times, with the middle term successively 
at (, 2, 3 and 4 -. 


6.83Mj^-*- Z7.S2 M, 4 --6.85 « — 2Sr4,400 f**Kidleferm tii I. 

6.83M, + 30.S 4 8.55 Mj = — 34 S, 200 .. .. z. 

8.3330.3Mj -r 6-83 ■ — 323,000 ~ - 3 . 


Tollcct terms in (1) and ( 4 ),divide 
each equation by the Coefficient 
of its first ierm^artd obtain the 


6.«3Mj-»- 27 . 52 M 4 6.83 M = - 210,000 

- 


4. J Simplified denvati ves (b). 

M. 4 

0.5 Mj =- 9.320 

(U ' 


The center Moments. 

M, + 

4.44M^ 4- 1.22 Mj c- 50,60 0 

a) 

r V 

Hominy found the 4 moment^ M, , 


+ 3^4 Mj 40.82 M 4 = - 38,80 0 

(3) 


Mj^, Mj and , wc newtind the 


Mj+2.oaM4=- 15,350 

W J 


Center moment of each side astallews; 

( 2 ) minus (l) 

3.94 4 1.22 M, =-41,280 



M = - 4.1950 ttibs. 

€ ^ ^ 

or 

M^ + 0.3lMj «- 10,450 

(5) 


M’awl-illlil*- * 4 2668 - •• 

(, 5 )minus( 5 ) 

3.53Mj + 0.82M,^=- 28,550 



® v ’■ 

=4 525 5 •• •• 

or 

Mj + 0,246^^*- S.S'OO 

C 6 ) 


« 8 2 

(4) minus ( 6 ) 

1.754 M 4 =- 6,850 



The required steel f>crfoo-t oicuiveri- 


M-« - 3,900 ft.lbs. 


is now tiqured tor the respective moments 


From(t) Mjs- 7.600 .. 

•• 


at Corne^'i and centers of all sides, from 


From (5) — 8,094- " 

•• 


^ , far M in fl.lbs,and da net- 


Fronr* ( 1 ) M, = — 5,150 •• 

- 


thickness af slab m inches. 


The Middle wall has no stress except as a column , but « nominal reinforcement is provided h> take 
CarC of temperature and a possible one sided water pressure . 


Dovrid/^.MoUtor 51/2/24. 


39 












































Art.29. StaficoKy Indctarmimai'e BcQrr?&ond Rigid Frames . 

Problem 29A. Beam fixed gfoKie gnd and simply supportgc/ af fhg oiher end. Sin gle l oad P 

There is one reduMdon+coMdi + Ion.Sec Probicm 6^ In ihe preieni analysis i-he and reaci-ion is 
treated asthe redundant and the pnnci pal system is thus a can+i laver beai'n tixec! at A . 

Thgqgngral moi-nent g O'^ ■ -i^ -for iha rnonien*- 
|P at an«y point distant i -f'm IS by £q I5B, 




■tm. 


-t 


■ -*j 


A,I—- 

+ 


- e 

Given Case. 

|P 

_ 


1Xb 




(I) 


M,area. 


fr- 

u 


TT 


t -- 


Mjj area. 


A^Jl 


Ip 

4^ 


area.. 


Th e elasticity equation for one I'edundantXjj/s 
by Eq.(l9C), whenthe supportatB is riqicl, 

. u) 

Firstdrovv the and the momenr di’aijt-amS 

, tor the load P and the Com/entionol Icod X|^=-|, 

o yr, 1,4^ tbof both a re positix'g tor the p'^esent" 

ftSSum ptions.mdi coted on the rgspeef ire 
^ T he two mfe^rals in £<}(2) are notr/ eroluoted 

jX^ uSin^^thc substitution tormu/as o-f Case 3,Iable26A . 


C I 

/ ~ tor the Icnefth a Common to both the and the oreas • 

and y^M^dK = ^ .£f for the len<)th 

These VC lues substituted into the above (2),furnish the redun<d«.nt Xj, as tol/ou/s : 

= Ab, i_ , end since El is constont ,then ^zf+bj^R^^which Jiscrte.lm E/j(/) 

^ives-—M,= P(x~bj-^LE a ^2£-rbj I <*s th e moment egunii on for anij point between .Aandm. 

When x = l), then M^= - bj; and when x = then M^- f-rbj = ^ b (£■«• b j. 

We Can now draw the complete M^diO^ram as shown, natin<j that the actual momenti arcotoppontesigns 
Problem 29 B. Beam fixed at one e nd andsimp/y Supported at the other, lintiorm Load pp erft. 

This problem IS eKaeily like Z9A excc'ptthe M^arca. 

^ r* X > . . 

Thetwo mTg<}ra is 1 are evaluated as per Cases Z9uS. 

j ^o^^dx 2 , -from Case S,Table 26/i . 

/^M^dx from Case 2 , Table ZC A . 

J 3 T 

The above £<^.( 2 ) 0^^0171 t^iVca the redundant Xj^ , as 


Vi 


Hoad pparlin.ft. 


2 

Given Case. 


p'per lln.ft. 


L4‘ 


area 


El 8 ~ El 3 ^ 

The rcactionot A = p2—X,=. 


Xi,= ^ = reaction at &. 

® a 


f- 

— 


Xb=-l‘‘'P 



I’-f' 


40 


Substitutimj the value X^ into 9Ives 

the moment at any point distant x from 8, as 

M, so when xa ^ and when x»f M »M» Pf.- P^_ 

* > * ^ X a "'S^' 

The equation M^=.^(x-^) enables us to draw the entire 

My moment diagram -for the actual Loadmq Shown 

and it should benoted asm Problem 29A , that the 

actual signs oi the moments are opposite To those found 

from Eqfl) because the abscissa x is really tahen maneqativc 

direct-ion.. 


40 


























Cli^lSOOP iw^ 



Problem 29C. ColumyrFixgd top and boHom, variable Mome.r\f of Inerha^hor. Load P. 

Simple beam as principal sysfem wi+h end moments av\d Xj, as redundanfs- 
Required to^ind ihe end ynomenH and at also the horizontal dcflccrcon <5^ at m . 

Data ■ 

1= \ SOOQ 111^=0.725 ft't ~ = 1.378 
1 = 100 0 in = o.o4-83f* . -L = 20-7 

p= 2 l kips, a = 6o', “bris', ^»7S'. 

'This problem involves only+wo 
redundonfs,because the colui-nn 
is tree to expand vertically. 

The Moment IJiaqrams 
torM^, M^and Mjj h ere Sho kv n, 
furnish the data torevoluatiruj 
the tivo redundanf momentstrom 
the closticity equotions I9C and 
M Is Ihcn found from Eq l98. 



L 


Mj^orca, 


The elasticity eguof ions for unyicldinq suppoi*fs are*. 


dx 


-0) 


Introd ucinqthe numerical integrals into Eqs(l) 


then 

42.0 X^+ “ <^76 P 

51.9 X^ -rSio.zXb = 2478 P 

or 

Xj^4- l•234• Xb = 14.05 P 


X^-e 5.980 Xi, =: 47-70 P 


4-746Xjj = 31.65 P 

Finally, 

X|,a6.67SP = 140.0 kip ff., and 

qive, M 



The evofuafion ofthe integrals isfr omCa$eiZ^3T-7CA. 

4/M M.dx=l:lZSa,.^,?a+2£Tb,^r^+n=247gP. 

ly'ob £ f a 0 L p J 

■^yM^M^dxs + SI.9 


ir/M^dx=.l:378arj + A^ hS ^_2o.7b, ^ ^ 42.0 

ly » "s i- e g*-i 3 

f M^dx = L17gAx£.^ + Z£:lk[£L^9^^\l=3lo.7 

ly b 3 e*- 3 Le^ e J 



-/164 


also draw the M^diaqram for P= I acting atm. Then • 

From Table 26 /i,Case 3 , evaluate the integral over the lengths a andb, 

nohnq thot the end ordinates of the M* diagram are -neqative moments. 

Then E^i^=l.378«^»| [2xU5.3-l64j + 20.7*^»k[2xl4S.3-l4-oJ=|4-6766 . 

and 5 ^ 14 876fc -. >4-8766 , 0.0344 ft. = 0 . 4 l 3 in. 

F 4-.320,OOO 

This same problem should be solved by area moments, Art.22^ 
observin<j fhat the end reactions of the elastic weights must be zero in 
the present proble m, because the end tonqent s to the elastic curve orezerOj 
qivinq ton ^3 = 0 , or R,= o , by Art.24. This condition always obtains 


M^diaqrom M^oreoforpsfixed end of any restrained beam 


41 





















































Probtgm 29 D. Column and Carttflever beam supporfin^^ loads shown .F»nd Moments 
Q<'/^. B ond C, also the horizonhal Ihrost' at/\ and C. Connections of-, BC arc n’qtd • 




>>=i.8kip& perft. 


'HT ^ 1=8000 ~ 

- 2*36' 


Uscl = 0,6 and I sl.O 

' ^ z 

(cl) Given Case. 


'P= 800 kip; 

1*10,000, hsU.'zS 




(1)) Condition X=o,M^area. 
Prin.System a 3-hinqed circh, 
with moment aPapplied at B. 




(e) ConditionX^=—I 


Moment* Eguotions Qfcording to £qd9B . 

M^= - M^X^ -for any point X. 

M. 


Elasticity E<}uations.unyield'mq Supports . 

^/m \4jc t- 21*>/m M,dx+2^^/M M dxts-^/M M dx 

ij ^ ly • b i/ “ c ly - - 


0 - I. X„ - o 

A 


o a> 


Mg*-Qp- O 

Mg= 0-0 


+ \.'X^ - 


•\ - 


— O r—Z4.8kipft. 

O = — 534 . 30 *^ the beam. 
0 c+ 1265.7 top of Col. 


=t/M. M, 

^yM^M^dx+^yw^Mj^dx+^yM^^dx=M^dx 


= 0 - 0 - o 


- I.X^ =-6 3 3.0 at base of Col. 


M, = -608-0.5X^4-0.5Xjj- o =-1-12.4atcent.of beam. 

H ^ - H^X„- H^X^= o -o-« Xcs_, 5 s kips. 

Notelhat 1600 kipff = the applied ext. mom. 

T o evaluate the above integrals we employ Table 26A,Cose 14,-forthe 3 integrals involving M^and first 
■Find at4 e^uaito^S 1460, ond at B e<)ual to Z 62660 ,osth« static moments of the M^areaoboutXandB. 
Theother irttei)rals are found by applying Case Z. -The 'foUowimj numerical values arc thus obfairted*. 


' /M^M.dx =--^[1x262660 +o] = % 120 

^y. *iv L J 0.8*36 

•? dx 3 . 4 .J., Po-r IX651460 1 ss + 6S'l46o -.*.22 620 

^4 0.8*36 ' 


ISX^- 7.SXg 


W.M^dx 

« 0 



M^M^dx 

= -l[ 

61. L 

* 0 ] 

6x0-8 


= 0 



M. M,dx 

_h_ 

.flxl - 40 ] 

12.25 

b c 

6r, 


6X1-0 

M^dX m. X X 1 
7 31. 

_ 36 
3*0.8 

= -1- 15.0 


i/Mgdx =e‘X I +Jl K I iLii =-1-19.083 

17 ^ 31, si^ 3 x 08 3*1 


1^-ZS = 44.083 
3*1 


— / M^dx = Jl 
iJ *■ 31, 

The 3X*$ came •ut-r3h*win^aisuinpti#ns(a)*vereCorrect. 

42 


Numerical Eauations . 

O =-9,120 
- 7.SX^-l-l9.083X^^-2.042Xg =4-22,620 
O - Z.oe2Xg4.4.083Xc= O 

Xg - O.S Xj,-= - 606 

-Xfl +2.S42 Xj,-0.272Xc=+30I 6 
2.O42Xj,-0.272X^=. Z,40& 
-Z.04ZXg44.083Xe= O 

3.811 Xc= 240 8 

)^*463 3.0 hiptt., X^*-t l26S,7ar\dX^4’24.8 kip-ft. 



























































91*0 = 


ProbI 6 in 29 E. Column Suppor'tcd on Three B&am^ over a Pitas shown infi<^u^"g- 

Find Bending Moments mfhg Beams and thg deflachon under the Column.. 

Th« 3b«ams are lo'x 12.4* I* M3.6 m‘^so.oo54S-ftf, SsZ2.72’in’ 
Th« Column wiih load P=i zo kips,is supported onihe beomAB^ 
and fkc beam A B is supported on thr pit wol/s of A oi*d B, and on 
two cross-bcams C and D, hfhich latter a^ord elastic supportsX,^, 
and X^, and these are freafedas redundants. The princi pal system 
is the beam A-B, Simply supported at A and 6- 

The elosticity Equations I^C -for 2 redundants, are ; 

S„=i./M^M^dx-i<>^/M^dA-ib/M^M^dn = losTXa 

El/ “ El/ ri/ 



-et 

fn this problem and ore not zero,and must be 

croluated in terms o-f the unknown loads X^^^and X|j,as the 

de-flections atthe centers otth^ cross-beoms as follows*. 

- i0.67X a.. and c jz - ZQ.Bi A b . 

® 48EI. ’ dfETi 

Since El occurs in each term of the two EqsO) and iS Constant^ 

we can cancel it from all terms. 

The intCdralsare noweyglug-ted from Cases 3Qnd4.Tablg Z6A . 

yM V1^d|{=l^(i%2.67)-r i^P(s.34r2) +-3P(2.fc7td^4-= 30.68P--{l) 

y"M„M,,d« = -r |[3 P(4- -t Z.67}-tZP(2 * S.34-)J-|-l«P(2-^2.6^=30.68P--^i) 

y'M^M dK=4iX£7^Z.67+ ±[z.S 7(2.67«.2.67; ■H.J}^.S3-^S.s4|rl:^»S.S4=24.«8-(2j 

y^M‘d>C*4 K I. n Z8A- ... ..<-4) 

/M*djc= 5. K ± » za.4 ... (5) 

y o 5 s 

Note here the t^juality between (i)and and between ^4)and.(Sj . 
The numerical equotions and fheir* solution Follow : 

30.68P-Z8.4X^- z4.88Xjj= |0.67Xal 
30.68 P-24.88 X^-.Z8.4 Xb=Z0.83X|^ 5 
39.07 + 24.88X^=30.Gap 

24.88Xq +49.Z3X^^= 30.68 P 
According to 19B,the M,g.9uotion is - X^ + o.637 X^, = o.78S P 

Mj,Xjj= M^-0.S7PM^-0.337PM(,. X,, + |.97 SXh« 1-285 P 

Hence M = ^P-2 x0.S7P-ZkO. 337P« I.I86P= 23,6Zki'pft. 1.33BX^,—O.4S0P orXjj=0.337PandX^0.57P. 

in ■ V . — - ■ T -i-r 

and stress'inbeamASis f=Mm - 21620jili = iz.Soo Ibs.p.s^j.in . 

* 5 *,X*72L 

Reaction A = fK- io-|.X„-iXb = o.i6 Kip. Reaction B=B^-B^X,^- io-^X^-iXj,=l.7Zkips. 

and ArB-tX^+Xj^aZO.ozkipSaP. The reactions C 5.7kips,and reactions D =^Xi,a3.37 kips . 

TheDetlecFion ^ sc— /^f7l<M-,dx and is obtained by integrating the area against theM^areaWPsl. 

Thus yivl,^ld ^dc= 4 aO.S4*4t |.[o-64(4r3;+Z3.62(zr6j]+^[z3.62(6rz)+6.88(3^4)]+^ xS.88k4« 163.57 

For E = 4, 320,000 kips per sg.ft. and I»0.00548 ftl, 4,3Z0,O00 xO.OO548 ~^’^^^'^'^'^°*^^^ 


El 

dx = ?£jliSb 


- 0 ) 


ihcipat system. 



P 


P 


M. Diagram 




> z -w- z 4' 
M Diagram. 


43 





















































Problem 29r. Three SforyBent, Rigid Reinforced Concret'e Frame. 

a/z 


l6Kjpa> 10' r T 



IW 1 S 


^\faO 


2b 

A B 

Given Case 



TTTTi 

Condition X=0 





. n 

I I 
\ 


Condition X=-l, 




Condition X,=-l- 

M, Area. 



We cut ihe bent into tu^o e^ua.1 hafv'es,obtamin<^ 2 Cantilever faosti for the 
principal system.The n qht half ii analysed for Moments Mj,€tc.,notiVuj 

that 4he moments for the left le^ will be eifual and of opposite si^n tothose found. 

The problem actually mvolres S redundants~a shear, a moment and a thrust at 
each of the beoms cut. TTie moments ore z.ero because for ee/ual end moments on 
the beams,the points of controtlexure are at the centers where the Section is token- 
The thrusts ore zero except at beams where horizontal loads are applied,as at the 
top^ And then we may assumethat aLoad as H is divided equally onihe two Cantilevers. 
Hence only three shears X,, Xj^ond mustbc evoluated as redundanfs. 

The verticol displacements 5, and of the points of application of ffie 

redundants in the direction of the X's are also zero^hence the elasticity 
became zero eifuations as follows*. Evaluation of the Inteejrals. 

XyiM.^dx-t Ay^iVJ, Mjdx t Mjdx =.^M„M,dx The intci^rals Con all be evaluated as per. 

Cased-, Table 2&A, for twotrapazolds ZandY 

^y^M^M.dx+^yM^dx + ^^yM^M,dx as /'^zMydx*! [z,(z>;tyj ^..r^(yr 4 2 yjj 

6 /m M.dx + ^i/M,M dx -t = Mdx 

» Lj 1. } ^ IJ o S 

The intcgrois ha/e thefollowinq Values : 

^yiVl^M^dx Jm,(zb,-rb) ■rmfbj+i'b)] = 4 -^ 0,553 

^jVnj2bjrb}+n'(b^+xb)] « tii9,47o 

.1. /M^Mjd*= ^3[^^j(2b3+b) + m(bj-r 2b)] = 4-129,600 

i= ^ [b^ ■rb,b4b*’]^ t Z,oo7 

1. yM,MjdK = ±yMjM,dx^A f 2,007 

i.y yMjMjdxn + t- 3,295 


ind 




-j j i^bVb,l>4b‘]4^ (o40-4b^]=. +• 2,4-93 

1 y .. t 3,521 

.L 4- i,\. 

I y » 51 ^ ■* 31. 


Numerical Equations. 

Z,99 3X, 4-2,007X^4 2, 007Xj = 60,533 
2,007 X, 4 3,521 Xj^+3,293Xj = 119,970 
2, 007X , 4- 3,293X^^44,10 3 X 3 = 12 9,6 00 
X, 4-0.806 X^ 4 0.806 Xj = 32.3 
X, 4-1-752 Xj.-^ 1.638 Xj= 59.5 
X, -^|.638Xj_4- 2 . 04-2 X^= 69.5 
0.996X2^40.832X3= 27.2 
-0.114 X j_40.4o9 X 3 = S-O 


4 , 103 


X^ 4- 0 . 88 X 3 = 28.7 
j^X 1 4- ^^59 X^=_ 93j8 

X,= »-764. X^-=<.»9.z8 4 , 42 X 3 = 7r.5-Xj=l64 


44 





























ProbI em 29F, Continued. 

Hov-m^ found the three redundant SKcorj X,=-»- 7.64 ki'pt, 14 .za kips,andXj=l 6 . 4 'kips, 
we now employ E<j, 19 B tofind+he moments -for various points of the principal system-forcer- 
tain hei^^hts >c above the <jround. Then tor the riqht hand post we find 

When x= hj ottop of post o -= - 82.0 kipft. 

- X=>,^ just above joint ^S.z .. .. 

justbelowjoint 54.7 .. .. 

- x= Ti Just above joint M J m, - b.X^-t). X,- ^^.9 - •• 

.• just belowjoint M, = m,-b.X-b, X^-b.Xj-xH b.(X. 4 -Xi^X 3 )=- Z 4.9 . « 

* x= o,at boss of post -bX,-bX^-bXj =K..h 3 _ b(X,-t-X^+X 58.5 « .. 

The moment on the top beam, nexttothe posts b^Xj = — 82 .o kipft 


ft ft 


■2 ?2^ 


=: = - 99.66 


Pa I kip. 


.bottom" .='b,X, = - 68 . 7 e ~ •• 

Ai a test, the sum of the moments above and be low a. joint, must c^jual the beammoment. 
Thus- s'll. 6 « = 45.2 + 54.7 nearly. AISO b.X^ = M'+ M, = 68.76 = 4 - 3 . 9 -t- 24.9 nearly. 

Problem ZSGr Find the horizontal Reflection at the top of the bent tor H^lbkips. 

The a«ljoinin<J diagram shows the moments at all points 
of the post, as just-found in the previous Problem 29 E .These 
moments cause the deflection of the post and each pos-t qcts 
)^fx horizontal lood. Hence the work I 4 d moy be used tofind 
the deflection. wil/ evaluate from Table 26 A, 

The Madia<jraiT» is drown for a horizontal force Pa I kip, 
and the inteijration is evaluated-from Case 4 ., in 3 fenqths of 20 ft. 
Thus: 

Eli, = m,)] -h ^ [40 (j m; - MJ* 10 (M.-- 2M,)j ^ 

M [io(2Mg-M, ) 4o(M,-2M,)]= ^|^6»*I4 + S87.^=20,I92 

Hence, i- 20 .92 



Dloqrom. . .^. 

I«l|300in^=O.S5S ft”*. 
E« 504,000 hips p.S^.f^ 
for Concrete. 


s: 0 . 0723 ft. = 0.867 in 


5a4;OOOXO.S55 
It should be noted that a fiqure like 



is merely atrapazoid with one Arid ondinai-^ ne^at'urc. 
iVhilc ifwas unnecessary to introduce the actual values for I ivx. solvin<j for moments 
in Problenn 20 E, yet when def le ctions are to be found if becomes imperative tb use the actual 
value of r, observin^^ also the units of length employed m solvincj the moment problem. 

It loads H had been applied ateach story ofthe bent, that would tnt.rely a.I-tev’ 
the Area in Problem Z9E,and add Slitjhtly to the numerical Work ot ■flndin<f ihf -f-lnal 
M,t d\ae\ram. 


45 
























Problem 29H. Seven Sfory Bent,subjected-fo honzonfal Loads . 

Assumpfions as for the three sfory bcnf, Problem 29 Constant I. 



-I- Xjy'M.M^a* + x^y'M.M^aK -r Xj^jvi.M^ax. t-x^yM^M^aK-rX^yfti.M^dxsylvf^M 
Xy^M^M,dK+X,yM*ax + Xj/M^Mjax -eX^yw^M^dx + Xj^M^M^dx +-X^yMjM^ax+XryMjM^axyNf^M.^ax . 
XyMjM.dxi-XjyMjM^dx+Xa/MjdK + X4yMjM^dx 4-Xj^M^Mjax + X4,yM3M^dx^-X7yi^,M7ax=yM,M,dx 
X,yM4M,dx4-XjyM^Mjax dx -♦- X5yM^Mjdx -rXtyM^M^ax+XyyM^MTaxeyM^M^ax 

XyMs M.dx +xyMyMidx +XyMyMjdx -I- dx -tX^yM^M^dx +X7yMsM,dx=yM^Mjdx 

X.yMtM.ax+xyM^M,ax7.X3yM^M,aK-»-X4yM^M4ax -eXs/M^Mjax +Y^^w\^d% 

xyM7M,ax4xyM^M,dx^-XjyM7Mjax+x4yM,M4ax +xyM, Mjdx+xyMyM^ax +xyM'a)c s/M.M^ax 
All iheie i-ntegrats are evaluated as p*r Case-4-,Table 26A , notmi^ thotthose consistin<^ of 
like subscripts asyiM^M^dx and yM^M^dx ^ are oi^uol by MoAvi/eII's Law. For the firsfy}w,M,dXj 

we qet = +hj -f hy ) 4-h4(mj+ 4 -hy^iTi^+m^^ . 

For e^ual story heii^hts this qiVes, after Colleciintf terms ^ yM^M.dx +-^ 2 7 e, 4 -zo . 

Similarly find all the aboire integrals era/uated as Poltows : 


yX((^Mdx= bhjifm + i ^ =-27 9,420 

y*^M^dx= bh|k*m4-£(ni,-nti,^a27S.7.l o 
yiM^Mjdx* bK |^*w-*-^(m^4m7)j=272,i4»l 
yM^M^dx=:b!h ^*m'^£(mj-<-m7)J= 254.438 
yMy-dx = bh w ■•-£(nv^ 7)J= 27 6,19 8 

yW.M^dx = bhj^<i-^ i:(»ns4m7)J=ri74,6o4 
yM^|y^dx=bh[i(»nfc-i-na7)3 = loz.Sifc 
y^M.^dx = b*^^^h-t = )2,<i72- 

yM,M^dK= » 10,368 

yw, IVl^dx* io'S.j h. - 8,640 

yM.M^ax^ = 6,91 z 

yM,Ms-ax= iTS-jh. 5,184 

.3.456 

yMM, = kH. 1.728 


yM^dx = b‘ lyjh-f- = 10,944 
yM^Mjdxrb'^^h ... * 8,640 

yM^M^dx* .. = 6,912 

yM^Mjdx^b^^jh... * 5,184 

yMjM^dxsrt^ ^Jh... = 3.456 
yjVljM.yax= b*” h . ... ^ 1.728 
y M^dx = b'‘[:S3h4-.|.j= 9,216 
y M^M^dx = b*^ h ... X 6,912 
yWjMydx^ b^'Sjh.... _ s.184 

yM^Mgax=b’-^Jh ... = 3,454 
/MjMydx = b^h.. 1,728 

~rh6if- numerical ralues inserted into ihc afoore seren 
elasticity eijuations, afford a solution iorfhc beam shears 
X,twX 7 . SaeTables 29 M and 29 J for the Solution. 

46 



7,488 

yM^Mydx = 

S.ISA- 


3,456 


1,728 

yWydx = 

5, 760 


3,456 


1,728 

fMldit. = 

4,032 

yM^M^dx = 

IJ'LZ 

/M^dx = 

2.304 


































































TAB'-E 29M. Solution or tmc clasticity cquations. 


OPERATIONS 

Coefficients OF THE X's 

MumenculliV’M 

X. 

Xx 

X3 

X4. 

Xs 

Xc 

Xr 

K 


I2>72 

10,368 

8,440 

6.912 

5,184 

3.456 

J.728 

2 79,42 0 


10,368 

10.944 

8,64 0 

6,912 

5.1 84 

3,456 

1.728 

27 8,2 1 0 

^[Eion 


8,50 0 

7.0 7 0 

5,660 

4.245 

2.829 

1.4-15 

228,800 

EWZ), 


2.444 

1. 570 

).,2S2 

939 

627 

3 1 3 

49,41 0 

E-<l(3) 

8.640 

8,440 

9.2 16 

6,912 

5.184 

3,4 56 

1.728 

272, 162 



7.070 

5,890 

4.7 IS 

3.535 

2.355 

1. 177 

190,500 

E<(3). 


1.570 

3.3 26 

2.197 

1,649 

1. 1 0 1 

55 1 

8 1. 662 

^LElW-l 



1, 007 

806 

604 

40 2 

20 t 

31, 700 

El(3)2 



2.3 19 

1. 39 1 

1.045 

699 

350 

49,962 

El (4) 

6.ei2 

6,912 

6.91 2 

7.488 

5,184 

3,456 

1,728 

256,438 

tH^CeiO)] 


5,QGO 

4.715 

3,778 

2,828 

1.685 

944 

1 52,i00 

Eh (4), 


1.252 

2,197 

3.710 

2.3 56 

I.S7 1 

784 

104,138 

ai^LE^(2).i 



804 

643 

482 

322 

160.5 

25,32 0 

Eq (4)z 



),39 1 

3.0 67 

1.8 74 

1.249 

623.5 

78,8 18 

^i?KEl(3)J 




837 

628 

420 

210.Z 

30,000 

E4 (4), 




2,230 

1.246 

8 29 

413.3 

48,818 

Eh IS) 

5, 184 

5,184 

5.184 

5.184 

S.7GO 

3,4 56 

i.728 

226,198 



4,245 

3.535 

2.8 28 

2.1 20 

1,413 

7o8 

114. 100 

E<(S), 


939 

1,649 

2,3 56 

3, 640 

2,043 

1,020 

\ 1 2,098 




604 

482 

361 

241 

1 20 

13.000 

Eq(S), 



1.045 

1.874 

3.279 

1.802 

9 oo 

93,098 





628 

471 

31 5 

157.6 

22.500 

Eq(S)3 




1. 246 

2 , 8 og 

1,487 

74 2.4 

70,59 8 

- 4 I^e-i('+W 


• 



696 

463 

230-8 

‘2-7,30 0 

Eq (5)4 



■ 


2,11 2 

1,0 24 

511-6 

43.298 

E«(S) 

3,4 S6 

3.4 54 

3,456 

3,4 56 

3,456 

4,0 32 

1,728 

176,604 

>^E«( 0 ] 


2.829 

2.355 

1.885 

1.4 13 

942 

472 

76,200 

Hq (6)i 


6 27 

1. 1 0 1 

1.57 1 

2.043 

3,090 

1,256 

1 00,404 




4o2 

322 

241 

1 6 1 

80.3 

12,670 

Eq (6)4 



699 

1.249 

J, 8 o 2 

2.929 

1,175.7 

87, 734 





420 

315 

21 0.2 

105.5 

15,0 80 

E^q (6)2 




829 

J.487 

2.7 18-8 

1,070.2 

72,654 

^iii{:ei(4)»] 





463 

3o4-0 

153.2 

1 8. 150 

Eq (6)4 





1. 024 

2.414.8 

9170 

54.504 

i^Eq£S) 4 ] 

• 





497-0 

248-0 

2 1,000 

(6)5 






1.917.8 

6 69.0 

33. S04 


47 






























































































TABLE 29H - CONTI NUED. 


OPERATIONS 

Coefficients of the X’s 

. 

NumcriCfllT*nr 

X. 

X2. 

X3 

X 4 

Xs 

X 

r 

X 7 

K 

PitT) 

1,72 8 

1,728 

1.72 8 

1,728 

1,728 

1.728 

2.304 

10 2,816 



),4 1 5 

I, / 7 7 

944 

I 08 

4-7 2 

236 

38,16 0 

Pqf?), 


3 1 3 

55/ 

784 

1.0 20 

1.256 

2 , 06 8 

64.656 




20 1 

140.5 

» 20 

80.3 

40.1 

6,330 




350 

423.S 

L) 

1 0 

1 0 

i 

1,175.7 

_ 

2.027-9 

5 8, 3 2 6 





210.2 

157.6 

105.5 

52.9 

7,55 0 

£‘>(7h 




413.3 

742-4 

1 . 070.2 

I.97S.0 

50.7 7 6 






230.8 

IS3.2 

76.4 

e,oso 

Eq ( 7)4 





511.6 

917.0 

1.898.4 

41.726 







248.0 

123.4] 

I O.'^IO 

E<i(7)s 






6 69.0 

1,775.0 

3 1.256 








233.2 

1 r, 680 

£<i(7L 







1,53 1.8 1 

I9.S76 


Kj = J2.75 kips. 


TABLE 29J.EVALUATION OFTHE X's BY SUCCESSIVE SUBSTITUTIONS INLAST5ET 




C 7 X 


7 A7 


CfeX *. - 


Cy Xg 




C j Xs 


CzXi 
Th< cocfficien+s 
C are-ft'omihe.laif 
5«t of Tables 29H 


X. 


102,8 16 


-^,400 
-22, 340 
- 19.280 


- 14,400 


- 10,020 


- 5.015 


-I0O4S5 

+ 2,361 

y, Aa.f-l 
17 28 


Xz 


64,654 


- 26,400 


- 25, 86 o 


- »6,2XO 


- I 1,390 


- 6,540 


- 3,198 


X, 


58,324 


-I 5, zoo 


-13,840 


- 10,030 


- 5,200 


-56,290 


+ 2.034 


-63,748 
4 9o8 

X = ^-^-<50 

-J.3SS 


50,776 


-25,20 0 


- 8,290 


- 47,330 
3,446 
X4--H3,3 8 . 3 s 


4J.724 


■ 24,20 0 


-II. 820 


-34,0 20 
4 5704 

y.5704 


Xc 


31.256 


- 22,600 


19.576 




5 S-II.C 


- 22.600 
8,654 

trM.I6 55.26 

I 

Values Safisfy 

Column MomtNTS- 


M, = bX, = 12X1. 36 5-^ -H 16.38ff.Ki'pS. 

Ms= iTJ, — bX, = 16 . 8 —16.38__ ^ 0.42 


Bfam moments. 


Roof 

12 X,^ 

16.38 ff.klps 

7tiJ Floor 

. 2 X^ = 

34-8 

6t^ .. 

12 Xj = 

69.7 

sfe - 

« 2 X,^ = 

120.4 

4tb 

IZXs = 

1 3 3.9 

3*21 ” 

12 X 6 = 

I5S.0 

2’lf' .• 

12 X 7 = 

153.0 


M'j = m,-b(X,-*-X^)= 16.8-S /.2 _ 

Mj b(X, + X^)=67.2-SJ.2_ 

M'j » m^-b(X,-#-X^-^Xj)=67-2-i2o.9 

M4. = ITIj- b(x, ♦Xj^-X,)=: 151.2-1 20.9 

= iTlj- b 2,‘*^X = IS/.2 -241.3 
Mg = = 2 68.8- 2413 

Mg :r h^fx = 268.8-375.3 

b^,^X ^ 420.0-375.3 
= Mg - b^fx =420.0-530.3 
My =m^-bS.f^X ST 604.8-530.3 
My = - b 2.yX =r 604 8 - 683.3 

M^=my- bS.^X = 823.2-483.3 


, -. 34.4 

■ 16.0 

r _ 53.7 

- 30.3 

- 9c.i 

-*■ 27 . s 

- 106.S 
-I- 44.7 

- 11 0.3 

+- 74.5 

- 7 8.5 

139.9 


48 






















































































































Problem 29 K- Find the Horizontal Dgf I ecfi on offhe top of f’he Seven Story Bent. TVoblem 29H . 

The M^orca represents+hc acfuol moments in the column^+hc deflection of which is to be found 

■for the top point’ m. The beom moments have been omi-tted. 
Since their effects ore alreody represented m the column 
moments. The problem is Solved in the manner illustrated 
m Problem 29(i^employm^ Mohr’s work E^uation^whence 
S = Jl M„dx . 

El ^ 

The diagram is drown for P= I kip , actin<^ horizontally 


*V|a>-4l< 

•>* J 

rlfc.4 l^« m 

*f4 

U)i=:l4 

, -S4d 

f*- J 5: 

’h 

a 

, -SS.7 

in J n d 

Vi 

, -^1 

t “M « 

V K 1 ->H 

— -- 

*(N 

w«^2.« 

\ 

-lot.5 

'-ao.s »t « 

A “*. 4 : 

\ In f *5 

'h 

-IIOS 

<-»7.S « "ii 

tv 

\ K A -IH 

n 

,-7«5 

i'js 

*£l 


\ 2 M K 

A 




atthc pe'inf- m , and the inte^jrntions are evaluated from 

Cose-d-, Toble 26^, in seven lencjfhs of 12 observm<] 

positive ai-id negative moments. The Values in'?” musf be 

Converted into E = d.^'ZO,000 kips per s<^.ft. 

The Volues _L ft.^ -for the three Column sections 
T 

Ore Kritten on the M,„ dia<^rom. 


M, area. 


area.. 

ITl 


The Substitution formula .Cased-^Table 7.GA, is 


r^andz^ ordinates moy be the moments , and the and ordinates are the moments. 
This i^ivesiKe seven volues as follovvs^in units of feet and kips. 


Henca 


7th story 

2! /m, m 
ly * " 

_ »2 »69‘751^4.0., „ ^ o.dvZxiaJ =.f- 28^800 

G Ih story 

•« 

_ l2x69.7S|l-^^^^7xl7+24-^4-l4^l2+-ZK2d-)j =— 96. 800 

sit? story 


^ ll«69.75jls3.7(zx24-+-36j4.30.S^Z4+2A26jlj =- 20^000 

d-lti story 


_ 12x28.1 jldO.I (2x36.^4«^4.27.s(36-»-2x4^J =-4o3 OOO 

3 ^ story 

•< 

_ 12X28.1 jIioA. 5 :^ 2 K 4.8 + 6o)4.4d.7(4^4-2x6cll| =-Sll.SroO 

"2*!^ Story 

#« 

_ I 2 X ir-o6|l||0.5^2x6o-«'72^-t 745/sot2x72)j =- 198,000 

1?! Story 

tf 

I2x l7o6r pg.s^2,t72t84-)4'139 9^72+2x8^ e=4-53S,O00 
Tofal = - 97a, So 0 

S =- 8 

70 S^OO 

- O.ZOf -ft. =r— in., indicating movementtothe 

fW 

£■ 

-4,320,000 


49 



































PROBLEM 29L:TW0-H!NGED ARCH WITH HORIZONTAL TIE . 



The principal System i» a simple span on t»vo su.pports,wiih X^sredundo 
The and stresses arc qiven on the left halt, and 5^ onthe ri^ht half. 


nf member ■ 




Mcmbei 


ight ho If of span. 
5^ STRESS DIAGRAM. 

Tie member 2L' 3 ^ 2 '/z»>«*-,^oft.lon<j. * 275. 

Table below gives 5«5^3l:=fi9032,^ S* 3^=1350 

TABULATED STRESSES S..SaAND Sx= S.~SgX^. ALL iH KiPS 
Ic. «il c*- . IgpoSsI .1 1 I* - 1 I -I 


Section 


Gross 

Area 

F 


Length 

on 


>P 


Sc5.i 




S.= S-SA 


O- 1 


40 


2L 2/i* Z* 


I.&2 


120 


74 


40 


1-3 


43 


l.oi 


45.fc 


1.12 


2L2;^i>2?^«%. 


1.80 


124 


72 


3270 


81 


- 2 


3-5 


74 


1.50 


III. 


2-25 


1.80 


I 24 


72 


8000 


IGI 


lO 


5-7 


93 


1.95 


181 


3.80 


1.80 


129 


72 


13000 


272 


10 


0-2 


56 


2L4«3* 5/IG 


4.18 


IG8 


40,3 


56 


2-4 


74 


0.42 31.1 


0.18 


2 L 4 x 5 * 5 /ic, 

I pH. IZx»/4- 


7.18 


129 


18 


558 


-56,2 


4-6 


93 


0S7 80.8 


0.76 


7.1 8 


IZ4 


18 


14-50 


14- 


-Sfc 


6-8 


98 


1.30 


1272 


1.69 


7.18 


.129 


18 


2Z40 


30.S 


-43 


I -Z 


16 


0.40 6.4 


0.|6 


2L2>5iK2K>ic 


1.62 


1 20 


74 


473 


M.8 


■r I 


3-4 


28 


0.40 11.2 


0.16 


2L3 k3k ’/4 


2.88 


1.20 


41.7 


468 


6.7 


- II. 


5-6 


»7 


040 6.8 


0.16 


2L2y2n2«ViC 


1.62 


1.20 


74 


503 


11.8 


2-3 


34 


0.45 


IS. 3 


0.20 


1.62 


138 


85 


I 300 


17 


14.4 


4-5 


20 


0.4S 


9.0 


0.20 


1.62 


138 


85 


76 S 


17 


0.4 


6-7 


0.4S 


3.1 


O-ZO 


2L 


).80 


138 


77 


239 


15.4 


- 12.1 


Sum 


32316 

3 z 3 ltt 


7-8 


62 


0.46 


595 


0.42 


2L 2^1*2* 


6412 

44I^ 


1.62 


120 


74 


4400 


68-0 


• 21.1 


69032 


1350.4 


50 




































































PROBLEM Z9~M.TWO-H<NGEO PARABOLIC ARCH RIB. CONSTANT El. 




Uniform load d oerft. 



The problem involves one redundant X, 
which represents the arch thrust,ond which 
when removed, converts the structure into a 
simple beam as the principal system. According) 
to Ec|. 19-C jthen for Co nstant EI and ne<^ lectin<^ 
axial thrust,the redundant X is "found from 
|.S»^ M^dx = O , giving the 

value X=r Mg dx ___/A 

yM^^dx 

With X known , £g.\9B givesthe momentat 
any point m as M^X-(2^ 

The M^Qrea for X=— I, is independent of 
the live loads and hence yM^dx is a constant 
depending only on the coordinates of the arch axis. 
From Table 26-A, case 6, find 

.(3J 

The c^uantlty depends onthe 

loads and will be evaluated forthefollowing Cases; 
For a single load P . Table Z6-A ,case 7, gives 

hence X= /MoMadx . 5 P^b fi-ab \_( 4 ) 

/Mi ax STTT'- 

For several loads P . the formula for X 
may be written as 

- 

For g uniform load p per ft . over the whole 

span , Table 2 G-A , Ca se 12, gives 

/*M- M dx as S. £ h £i at-L pT 
J o ^ 15 ^ 6 15 ^ 

hence X* -- i&) 

8h 

Fora uniform load over a di stance 2a . 
it IS best to evaluate J'^o M^^dx numerically 
by Case 7» using an approximate triangular M^area. 

When X is found for any given Case 
of loading , the n M^^^X . 


Temperature effect . The horizontal thrust , due to a rise in temperat¬ 
ure of t” F., is found as follows : For a coeff Icient of exponsion £., the change 
in the length 6 is moking the temperature thrust ~ 

tte , IS El ft_ (j) 


X4.= 




and the moment^at any axial point m, is M^^=-M^Xt. 

For an exhaustive treatment of the subject of arches, See the author’s 
Kinetic Theory of Engineering Struetures , 19U . 


51 
































































PROBLEM 29-N.TWO-HINGED RECTANGULAR FRAME.LOAD W ON POST 



B 


"n 

; [ 

1 

^ h 

c 

i 



+ 

1 

h-*i 


A f-I B 


\ -t 

'' •*•0 

-r / 

V__ 

m 


f V 

PRINCIPAL SYSTEM 


1 \ 

-I 


r \ 

Simple beam 


Ld\ 

+ l C 

/x=-i 




0 

* C n 

0 


M^area forX=-I 



As In Problenri < 3 cncral formulas 

for + hc Case of one rcdundanL condifion are 

V.i/MoJiMjL and IV1 = M-M^X. 

Using the substiLufion formulas 1 ondZ.Table 26-A, 

1 /M^ d* =2iL + iil = T? r2iL+ Xl_ 

V. . . 5T 1. I.J , 

which IS independent of the loading and applies to 
all two-hinqed rectangular frames. 


forX»0. 

According to the formulas 2, 3 ond 4-, Table Z6-A, find 

= J^x2r+J^[wr(2r+h) * Wr(r+zh)]+^.3h = |y-|%lj|rj^. 

L. y, htU ^ ° 




riJ-t* 1 

ii| 


1 '1 



The moment formula gires M^=M^-MgX.^Wr-rX , also M^=Wr-hX and Mg=-hX, 
The final diagram can now be drawn to show the moment aterery point. 

The reactions are W-X , X , — D = -eC = iC|p* • 

The change due to fernpcra+ure inthc length £ is 21£, for a rise of t^F., 

ttf £tfEI 


giving 




•*■0 • 

SPECIAL CASE WHEN W ISAPPllfO AT A. then r=h and the above formulas 
become x . IKi- hg-f. yi/, W gnd M.= h (W-X) , hX D=+C= Wb . 

For the following Cases of loading onthe two-hinged rectangular 
frame, the integral ^ retains this form and will 

not be evalua+ed again except for frames of different shape. 

However, in describing the several cases which follow, the and 
diagrams will he shown for purposes of illustration. 


52 







































PROBLEM 29 - 0 .Two-h\nged rectansolar frame.uniform post load. 



O' ^ / 

With substitution formulas 3 and tO,Table 2fi-A , evaluate the integral 

1/'M M d)c = -§.ln + *3^= B^Lfsh ^ e ] ^hich together with 

I; ® c 81 3 izl„ 4 I^J ^ 

the value = Vi f^h ., ^ .1, J,as found in Problem 29-N , gives 

>, rlt'^ f. "I'ph rshlo+eltl ph 

o o 

From the moment find M — MgX= pr^h-.C^- rX , — hX , Mg*0-hX. 

The reactions ore pli -X , X , and -Rs -»-R = hIi . Also Xi- -r >^ ^ ^ ^ ^ 4,=— M X^. 

^ Tt ^ iT^h^-U] 

PROBLEM 29-P.TWO-hinged RECTANGULAR FRAME . LOAD P ON BEAM. 



With substitution formulas 5,Table 26-A, evaluate the integral 

find X= T b 'T"- —T^rr—^o*' several loads P find X« - 

The moment Ec^. 19 B gives M^ss M^— M^X = — UX^ also . 

The reactions are X, R^c: ^ and R^^*^ , with X^ and as above. 

PROBLEM 29-Q. UNI FORM LOAD P pe r ft. over the beam. TheM^Area becomes 

porobola with middle ordinate ^ ^ givina .1 dx =-^*zh*^ and 

P j _ making M = _ hX , M-» M,» —hX ,ond R - Rj = Ef . 

4h(sfl.1-3e) ^ 8 • A » C d 2 


a 
X = 


53 






































































PROBLEM 29 -R.TwO>H»NGEP RECTAN6ULAR FRAME.CRANE LOAD ON ONE COLUMN. 



With suLstitution formula 4,Table 26-A, evaluate the inteqral 

h.3Pc]+ ^^[h.2Pc*hPc]=Pc[Jl^% ^ ]. 

Using the value ' -»• found in Problem Z0-N, toobtain 




riT-r^ kli 
TIT 


Pc-. (K'‘-r^)l. + hfI I pr 

^(IFttJ ei) J 

From the moment E!cl. 19*B. find M =0-rX below, and M sRc-rX obove the bracket. 
Also Pc-hX, Mg=—KXThe reactions are X , « P,and 

PROBLEM 29-S.Two-HlNGEP RECTANGULAR FRAME. CRANE LOADS BQTHCOLS. 




A ^ B 


-r 

±i' ’’tir 

-k 

0 

rn 

Principal system 
simple beam. 

s a 2 

C 

r 

• 

? 

P 


M^arba,X*o. 



Same AS Problem79-R. 


REOUNOANT FORCFX. 

with formulas ) and 3, Table 26-A, evaluate the inteqral 

z 


M^area fork*-I. 


^jM^M^dx= ^i^jr*3PcH-hxsPc]-h 


A?> 


X = 


■(LfXV ^ 

Pc = 




Lh^(|hi,-ei) J 


Pc. 


The mo me nt Eg. |e 3 qi vcs M =0—rX below.and M » Pc-rX above the bracket. 

» •' fn ' m 

Also Mg = Pc-hX . The reactions are X.and R = = P. 


54 
































































PROBLEM e9~T. VARIOUS TWO-Hm&ED FRAMES . 


The M^areas for various loadingt^for+he ca^es here ^iven^wiil he similar 
to those drawn for Problems 29-N to 29-5,ond the evaluation of^M^M^dx may 
be left fo the reader for practice.The integral_/M*dK will however, be i^iven 
here and each formula remains unchanged for the same frame,irrespective of th«loads. 



^ M^dx* giving 


1 /■ c 


Also - (V1^)C and M 




3.ttCE 





55 







































PROBLEM 29-U.TW0-ti1NGED.UN5YMMETRlC FRAME. 



- Cose 10 


yVI^M^dx ^ ^k , 6.2S A [ 5 ( 6 . 25 *25.4_ 5.3* »7. I^+6*24.4(6.25+5 3) + 6.2S*I7.I +S. 3 X 25 . 4 I 

__^ •- -J 


Case 3 

+ 1 *5.3* 2 K n.i = 1,871 
® -Cose 2 


— Case2_^—“•“*-F- Cas,ei --^*Ccis«2 —► r M/4v 

/M^dx = ^.6.2 sV |.[6.2s\6.J5.M..r3j+|.*53'=4.S6.X=^^|^\^.4.nl<;p, 

M^=M^—M^X = 1^8.29-4.11x3.125 = -1-5.45, M^=24.4 - 4-. 11 *. 5,725 = + 0.90 f+.kips. 

M* = 8.55 -4.11 X 2.65 =—2.35, M^ = 25.4 - 4.11 *.6.25 =-0.30 , M =I7.J-4. II* 5.3 = -4.7 


Reactions .Thg loads onihe span are first combined Into a resultant R=7.S3kips 
and the vertical reaction at C is Computed as Y=.7.53x^^5 = -2.S3kips. The reaction 
n)jisthen found from the condition that R,\( and must intersect in a point, giving 
its lever arm as 15.i ft. and making D^=Zi^^lLL - 5.5 kips. R^ is finally found as 
the resultant of D^and X,and R^is the resultant of \(.QndX. Rj and R areiound 
graphically after X is Computed as obove. 


56 



































PROBLEM 29-V. RIGID RECTANGULAR FRAME.THREE REDUNDANTS. 


w A B 



The general Et^s.lS-Capplicd+o the present problem: 



The numerical elasticity actuations thus become : 
14 G7 X,+ 90.0 Xj +-90.0 X^ 14 67 
90 X,4 12.67 Xj i- l.333X^ = I 53.3 
90 X,4 l.333Xj4 l2.67Xp * 26.67 

The solution ^ives X^I.OO, X^aS.SS and X^=—5.58 
The moment at any point mis given by Eg.l9B os 

M„- M„ - M,X.- MjXj - M^X^ 

The tour corner moments thus become ; 
=:WVl - hxl - 1k5.58 -t- 0<s.58 = + 4.42 tt.kips. 
Mg = O — h«l— O + lx 5.58= -4.42 •• •• 

o-O — Xj— O = - 5.S8 •• •• 

0-0- 0 +-lxXj.=-»- 5.58 - .. 

For c^ifferent loads on similar three sided 
rigid frames, only the three last integrals must 
be evaluated for each area . As may be seen 
from ProbI em 29-U, uriy combi nation of simuItaneous 
loads may be solved as one problem, merely by 
including all such loads in the Mpmoment area. 


57 





























PROBLEM 29-W. DOUBLE RECTANGULAR FRAME:. H\NGED ATBOTTOM . 




The complete solu-tion oi the above problem i& suggested for practice. 

The principal system^ a sim pie beam 4— 6 , was selected as offerini^the easicstsolutlon. 
Employing the numerical values,asgiven in the above sketch of the g iven case , the 
nine integrals-for the solution of the three elasticity equations 19-C, may be evaluated 
bymeansofthe substitution formulas Tabic 26-A . Thenumerlcol equations when 
solved, should give the following values: X^= 1.14-5 , X^—0.419, and 0.703 kips. 

with these values of the X’s find the moments at various points of the 
frame according to the general equation 1 ^* 8 ,thus-—, 
after which the moment diagram maybe drawn as shown. 

The reactions arc obtainable from R=: R^X^— ^^ 5 X 5 , noting that" 

the vertical reactions of the and Mg diagrams arc all z.ero. The vertical reactions 
R^,Rgand R^dependon P.WondX and the ir sum must equal P. The sum of the horizontal 
reactions must equal W. 




















































PROBLEM e9-X. FIXED PARABOLIC ARCH. CONSTANT El. 






M^ARCA,Xa-I.O 
-♦■momenta produce tension on 
Ihc lower side of the orch. 


Solution for a concentrated Load R 

The evciluation of 9 integrals by Table 26-A. 
/M>x = 

/M*d*=?r5 , 

These values substituted into Eqs.l9-C,give the 
following three elasticity eguatlons for Constant El; 

U,x.* ^X^4- Afl?X^=B^(, + £^). 

These equations arc simplified to obtain: 

x^t zKX^ =f|k{i+|.) -x 

X„f2Xj,+ 2V.X^ = P|b(l+a) 1.(,) 

X.+ |hX^ =E|b ) 

The solution of Egs.CO gives : 



For several loads 1^, etc., Co leu late the values 
in Egs-(2) for each lood seporately and Sum the 
results respectively for X^. ,X^^, andX^. 

Finally from F<|. 19 B, o btoin f he moment 
at ony point m of the arch axis as * 

M.- M„X,- M^X^- .-.(3) 

where , M. ond M, are the ordinates 

W • • • Q C 

in the respective moment areas for the point m, 
which may beany axial point,not necessarily the 
loaded point. 


For g uniform load p per ft - over the whole span ,the redundants become 
X , and X = X = 0« giving a zero moment for oil oxial points,andthus prov- 

cgpj b 

ing that a parabolic arch is most economical. 

The above formulas neglect axial thrust ,as may be seen by referring to E(^. 14-E, 

/f ^ 

ond this affects Xg when the arch becomes very flat. See Kinetic Theory of Engineering 
Structures^for full development of methods for bridges with moving loads. 


59 












































FORMULAS FOR PARABOU\C AND SFMl-ClRCULAR ARCHES. 



TWO-HINGED ARCHES 
P 

Etfuaiipnofparobola p— 
origin at A . ' 

y = 4-abil • 

^ V 


l\ H=£^fl+ab') 

8 eh I 

M^=E2b«Hy. ^ 
For load Pat crown ^ 

H =0.195^ M^ = ^-Hh. 



Equation of circle 

origin at center 


= 0.055 P0. 


Uniform load p per ft. 



with load Pat m. 
5ee Problem 29M. 


’ 

8h 

® IVI =Bfa-.Pa:-Hv 

m 2 2 ' 

=iP£i’-Hy=0 ' 

M = ei-Hh:.o 
c 8 


zR For Load aicrown^R 
H=0.3»8 P 
M^.=0.I82PR 
8^=0,019 

Uniform load p per ft 


H= Esh 

TTR^ 




^-Hy 


M^=Pa_HR 

2 

with load Pat m. 



H = 0.4-4-I8 pR 

pRjB 

S = 0.006 . 

El 


FIXED END ARCHES. 



H= 0.4-70 P 


M=M=-o.m8PR 

a b 


M^=-»-0.I48PR 


PR* 

El ■ 


Uniform load p perft. 



B '^o=M,= o 

'^m=Mo„rRy=° 


The parabolic axis isthe most economical shape 
for uniform loads, because M^o for all points 
on the arch axis. 


Uniform load p per ff. 

! c 



H= 0.56 pR 

> I 

I M=Mj=-o.io 6 pR* 
M^= 0.046 pR^ 

8> =0.005 

^ El 


60 



























































FORMULAS FOR CmCULAR PIPES . 

Loads in 1 bs.per inch of pipe. Dimensions inches. Positive moments-tension on Inside. 

rSurfgce of liquid 

Licjuid vulbs.percu.inch. 

C ^a= ~ 5 - p) 

M. =-o.5R^(>R-p) 
pipe. / ° 

^ ^ / M^=+0.571 R(^u;R-p) 

^-6sin8-i|^wR-pjR^. 
EMPTY FLOATING CYLINDER . 



M =Mu= +0.3I8PR 

ci b 

= - 0.182 PR 

Hor. di a. = 2 R+0.14- 
Y« rt.d«a= 2 R -O.IA9 PR* 
Pipe supported at A. 
CONCENTRATED LOAD ON TOP . 


Ma*M^=+4:iR". 

H or. dia. =i 2 R + 

€EI 

Vert.dia.= 2R- 
Supported on DAC.^ 
UNIFORM LOAD ALBS perHOR.iNCH . 


M^.l.spR" 
M^=:0.5pR^ 
M^=-o.64pR^ 
Support A= apirR 
Tfit M^pI^(l-^-0sin0j 




Hb t 

MkT 





P 

R 




Empty 
pipe 

■1^ 


Supported at A. 

LOAD DEAD WElGHTOFPlPE.pibs.S<^.in . 


M = 0.75 u) 


N! = M^=+0.137u»R* 
a b 

M^=-0.149 VO R* 


PIPE FLOWING FULL-SUPPORTEDON EARTH . 

Wi^t.of pipe=p Ibs.p.sc^in. 

0 Licjuid w lbs. p.cu.in. 

Total weights: 

LMj. Q=2f>irR+wirR^lb5. 

per Tof pipe. 




M. = 0.25w R* 

_b 

M s-O.siwR’ 

C <= 

Support As. wlTRt 


= - Mj- ^ ( 5 ^ 2 ^ - o.ojs). 
fv1^= QRjesine^-iCose-J ■»-^(Cos0-5)Jbet.A%B. 

[-0)sin0-|.CoS0-:J_^(Cbs0-r|:)l AtoD. 


2-T-1tV-4d 

,, ^a= - Q^- M*x M when 0tQn0s^+ ^ . 

V V When ^=0.04, then M^s-M^=-o.ooasQR^ 

u) = o.0 36i-lbs.cu.in. M^=+o.oiQR. max M sO.oiQRfor0=6i-39 

for water ^ 


PI PE FLOWING FULL-SUPPORTED AT A. 


PI PE FLOWING FULL- SUPPORTED ATA^VF. 

61 









































PROBLEM 29-Y STRESSES IN KNEE-BRACES . 

Knec-braccs may be introduced into afromc to replace rigid corner 
connections, tKu6 stif-Fening the Co nnecLions and reducing deflections. 

The corner moments of the f^ramc arc determined as above shown, 
after which the knee-braces are introduceo and the stresses produced therein 
are derived from the corner moments as shown in the figures below. 




The stress in knee-brace B. B, « M* <^'- p<>»itiye moment. 


”2 T ^compreeeion forne 9 otive moment. 

The modified corner moments are shown as shaded areas for the 
cases of positive and negative moments. 

The Sign convention e mployed in Problems 2SA to 29X maybe stated osfollotvs: 
A positive moment acting an a beam . produces tension on the lower side. 

A positive moment aciincf on the Corner of a r>gid frame ^ Figs. 29-X ,produce5 tension 
on the inside of the corner. The some rule appliesto on in-terior column connection 
as at point 2, Problem 29-W, noting that the tension dementis onthe inside with 
the positive moment. 


62 























































































































Chapter 7 


Wind Stresses in 


Building Frames 

Art3Q . The .Nature of th e Problem. In Art. 29, a nujTiber of 
problems 129 S’, 2 9 C, iiW. and' 2'9K) were solved which would 
indicate, to some extent, the innumerable obstacles to be 
overcome in analysing a large multiple story building frame. 
But before attempting an analysis of such a frame, a design 
must have been produced on the basis of certain assumptions 
so that interest is centered primarily on methods of rational 
and economic designing, rather than on methods of ascertain¬ 
ing the possible stresses induced by wind forces in a given 
design, which may have been based on erroneous assumptions. 

The numerous assumptions necessary to bring this 
problem within the realm of possible solution are so grossly 
approximate that any attempted refinement must be excluded 
on the grounds of imparting fictitious accuracy to something 
which is nevertheless unknowable. But, since the fundamental 
requirements of economy and simplicity governing the design 
of building frames are quite well established, it would appear 
futile to design on any other lines. It would also be wasted 
energy to apply some so-called accurate, but laborious method 
of testing a poorly conceived design, merely to discover some¬ 
thing vidiich was knov/able in advance. 

The value and scope of application of all methods 
of analysis, based on the theory of elasticity as applied to 
a strictly engineering structure, is in no wise questioned. 
However, a steel building may be classed as an engineering 
structure only when the frame is bare; but when clothed with 
architectural coverings of concrete, stone and brick, v;ith 
steel and concrete floors, tile partitions, etc., it becomes 
a composite structure, the nature of which cannot be appraised 
in terms of mathematics. 

Art. ’dlT 

of the Assumptions. In dealing with the bare 
the fundamental assumptions on which are based 
the so-called exact methods of stress analysis, may be 
evaluated, approximately, as follows:- 

1. The connections between columns and beams or 
girders, are perfectly rigid. This may influence the result¬ 
ing stresses by 30 to 50$^ as indicated by strain-gage 
measurements on riveted connections which are more or less 
flexible. 


steel frame 


2. The changes in lengths of members due to direct 
stress are negligible. This may involve errors of perhaps 
1 percent. 

3o The lengths of the members are the distances 
between intersections of their neutral axes. This may affect 
the stresses by from 5 to 10 percent. 


63 









4. The deflection of a nemher due to internal 
shear is negligible. Deflections in beams are usually about 
10$j greater when web shear is considered. 

5, The wind load is resisted entirely by the steel, 
frame. If this were true, tall buildings would sway in the 
wind three to four times as much as they actually do. In 
other words, the architectural clothing, and not the steel 
frame alone, is responsible for the rigidity found by 
experience to be manifested in present day tall buildings, 

7/ere it not for the inherent rigidity of the architectural 
clothing of the steel frame, this ,assumption would be 
seriously deficient. Fortunately., the combination of a 
flexible steel frame, des’igned for certain wind loads and 
unit stresses, with a comparatively rigid fill material 
like concrete, does produce tall buildings with tenantable 
rigidity, a fact which has been demonstrated by successive 
development in building construction to greater and greater 
heights. 

By the method illustrated in Problem 29 K, the 
horizontal deflection was computed for a 40 story building bent 
100 ft. wide by 467 ft, high, with seven columns spaced 
nearly equi-distant, The frame was designed for 15 lb, 
per sq, ft. wind pressure and unit stress of 24,000 lbs 
per sq, in, in bending. The bent carried wind over a width 
of 27 ft. The deflection for the bare frame carrying the 
full wind load, was found to be 6,5 in. 

If such a building, in its completed state, was to 
sway as much as 1 in. in the wind, it might border on 
financial failure f^om the standpoint of occupancy. 

There is, however, one mitigating circumstance which 
■ay make these vibratory disturbances more tolerable, and 
that is the period of the vibration which may vary from 5 
to 10 seconds. 

An approximate idea is thus given, of the 
proportion of wind load carried by the steel frame and- that 
carried by the fill material even though the steel frame was 
designed to carry the entire wind load. 

However, this is the price to be paid to obtain 
the necessary rigidity for a tall building, as rigidity 
constitutes an important asset to the owner, and the amount 
lirtiich the steel frame alone contributes to this factor, 
despite the severe load assumption, is in fact relatively 
insufficient to supply the necessity. 

The assumption that the steel frame resists the 
entire wind load has produced certain empirical results in 
modern buildings which must be carefully weighed before 
launching out on nev; lines of design or new assumptions as to 
wind loads and allowable unit stresses. Thus, silicon steel 
stressed to 20,000 lbs. per sq. in., would contribute 25^b 
greater deflection than carbon steel at 24,000 lbs. unit stress 


64 


In the face of theae facts, wl-ut easeful purpose 
can be served by accurate methods of stress analysis? They 
may appear pretentious enough; but desi^jners are not Justi¬ 
fied in claiming for them any extraordinai*y degree of 
accuracy when applied to a tall building frame. 

Art. 3 2. _Stress Analysis of a Given Design. Since ty.e 
design of any given building frai'ie is" necKSsarily accomplished 
by some approximate method based on more or. le ss artjitrai’y 
assumptions, the stress ai; ''ysis of such a design requires 
for its solution the application of the theory of elasticity, 
whether in the form of slope deflections, area moments, or 
work equations, from which the redundants may be evaluated, 
with the redundants determined from elasticity equations, the 
various moments and stresses become known. The points of 
contraflexure are thus fixed, and the analysis locates them 
as a consequence of the design. 

If the design is based on certain preconceived 
rational assumptions, the resulting moments will natura;j.ly 
be consistent with those assumptions, but not otherv/ise . 

A design based on a certain set of assumptions 
cannot be reviewed or checked by any approximate method 
based on other assumptions. Hence, the assumptions for 
any given design should always be stated, as otherwise 
no satisfactory numerical check is possible. Thus, suppose 
a building frame was designed by Fleming's Method No. 1, and 
in the absence of this information someone attempts to dieck 
the design by Fleming »s Method No. 3, Tliis would result in 
grossly misleading contradictions without revealing any 
actual errors. 

A laboriously conducted analysis of a poorly 
conceived deaign would merely lead to the discovery that a 
redesign was necessary and the latter would tiave to be 
tested by a repetition of the same laborious analysis. 

It is thus apparent that a rational method of 
approximate design is of the utmost importance, and that it 
is futile to attempt a so-called accurate analysis of a 
tall building frame because, in practice the result is 
probably no more trustworthy than that obtainable by a 
simple, rational, approximate method, sucii as given in 
Art* 32. 

Art. 33. Approximate Method of Design, With regard to 
methods of design, the' objective aimed'at, in terms of 
economy and constructive simplicity, can be clearly 
stipulated. Certainly, such aims can and should be 
embodied in the design with advantage over every other 
disposition that may be made. The only feasible method of 
design is one based on rational assumptions, economic 
principles, and enduring satisfaction as proved by actual 
results in practice. 


65 






The important stresses in a "building frame are as 

follows: 

i*. For columns, direct axial and bending stresses due to 
lateral wind pressure, combined with dead and live load 
•axial stresses, 

2, For beams, bending stresses due to lateral wind pressure 
alone or combined with those due to dead and live floor 
loads. Also end shears on beams, due to wind alone, or 
combined with those due to dead and live floor loads. 

If the points of contraflexure in the beams and 
columns can be approximately located, then all the critical 
wind stresses in any given frame are readily found for 
purposes of design without resorting to methods based on 
the theory of elasticity. The theory of elasticity is 
used essentially to locate these points of contraflexure, 
but if their location is already approximately known, then 
this laborious operation can be dispensed with, 

Hhe maximum economy in design for columns and 
beams, with respect to bending stresses, is attained when 
the points of contraflexure in the columns are at mid-story 
heights, and the-points of contraflexure in the beams are 
at mid-points of spans, 7/ith respect to direct axial 
stresses in the columns, the maximum economy obtains when 
all or most of the direct stress is carried by the outside 
columns, and little or none by the interior columns, a 
condition which may be attained by an appropriate division 
of the total wind shear among the several columns at any 
floor* 


Investigations made along these lines, prove 
these statements to be true for bents of equal spans and for 
all exoept the top and bottom stories of a frame. The top 
story presents no wind problem and the bottom story can be 
separately treated if greater accuracy is desirable by a 
method given in Art. 34. 

When the.ppans of a bent are unequal, a further 
condition must be imposed, if simplicity in design is to 
be preserved. This is accomplished by assuming the bending 
moments at the ends of all beams in the same floor level 
as equal, allowing the beam shears to vary inversely as 
the span lengths. 

The solution of any frame is now definitely 
fixed so long as the foregoing principles are not violated. 
The following steps will solve the problem for any given 
floor level, as illustrated on Fig. 33 A and described as 
follows: 


1. Divide the total wind shear acting at any 
story, so that the two outside columns get equal amounts, 
one-half as'great as for the interior columns. Then for n 
columns and a total horizontal shear W , each outside column 
will get a shear Hs V^— , and each interior column will 

get a shear of zw . 


66 


2. The beam shear in the outside span is then 
found from the shears in the outside column Immediately 
above and below the beam in question, according to the 
condition that the beam moment must equal the sum of the 
two column moments• Since all beam moments at the same 
floor level are to be equal, then for unequal spans the 
beam shears will vary inversely as these span lengths. 

Fig. 33 A and the accompanying data give a 
complete outline of this method based on the assumptions 
first published by Albert Smith, M. am. Soc. C. E. in 
the Journal of the Western Soc. of Engrs. Vol. XX, No. 4, 
P. 341, 1915. 

The assumptions on which this approximate 
method is based, are not seriously in error except in 
the bottom or first story, and in this case, the only 
modification that might be desirable would be to select 
a more appropriate point of contraflexure in the first 
story, outitide column, retaining all other assumptions 
as for the stories above. Thus, the moment at the base 
of the outside first story column might be made equal 
+0 m = o. 7 hH instead of o.shH. 

The approximate method of design by which points 
of contraflexure are assumed, yield moments at the Joints 
which may be developed by appropriate details for 
connections, and the author maintains that such a 
structure must function in accordance with the 
disposition of metal provided in the design. There is 
nothing seriously in error with this procedure, and it 
is the only method available. 

In the event that the structure is capable of 
offering sufficient internal work of resistance (with¬ 
out being stressed beyond the allowable limits) to 
balance the externally applied work, then the design 
must be regarded as adequate. This statement remains 
true even if certain parts are overstressed, provided 
that certain other parts are sufficiently understressed, 
in which case a readjustment takes place between the 
several redundants and the principal members, resulting 
in a slight permanent deformation of the overstressed 
connections. After such readjustment is once establish¬ 
ed by a single instance of maximum over-stress in 
certain parts, the structure will function normally 
thereafter without again overstressing these same 
parts. 


67 


WIND STRESSES )N BUILDING FRAMES . 

A 55 U M PTI ON S - 

1. Columns have points of confraflexuvc ofmid-sfory heights. 

2 . Beams have points of contrnfIexure of mid spans, 

3. Wind Loads are divided so that interior Columns fake equal shears of fivlce the amount 
taken by each exte r i o r column. All bcom moments of same floor arc equal. 

V. 



V, 

v! ^ t' 


1 ^. L 



Definitions. 



A B C 

FIG.33A. SECTION OF ONE BENT. 


CAiswind load at o floor. 
Wstotol wind load from 
roof downiro a certain story. 
H =hor. wind sheer on an exf. 
colurnn ai mid-story height, 

V = Vert. Wind shearafeenfer 
of any beam* 

P=axlol w’ind load on an 
exterior column. 

'*M3 ^ 

b= span of a beam. 

Hs - 

n ^number of cols, in beirt. 

All loads and forces in kipS, 
*^^5 AII dimensions m feet. 

^ 4 - Column moments m^ond 
beam moments M in kipft. 
Numberfloors from roof down. 


Tobulote Computoti onS m following order, wrltmq Values on an out | me diogrom of the fro me , 

1. Find wind loads tv;, u>, , ,etc. acting at roof ond the several floors. 

2 . Find total wind loads W,= ur; ; > W^=ur + 

3. Find hor. wind shears H m one cxt.Column. — ; H - . N - -.etc. 

' a(n-l) » z x(n-0 ’ 3 a(n-i) 

4. Find vert, shears V in outside span bos V,= H^i. [H^-h.t-Hj-h,]-, etc. 

5. Find beam moments - M=Vi . M sV . M, = Y, b • etc. eguniforall beams of Samefloor. 

I *2' 2 * 2 ^ ^ ^"5r' 

€. Find outside Column moments, m - m - Hiilx , m - HlJls. efc 

) I ^ > a— ^ , ”' 3 - 2 , j 

7 . Moke inside column moments double those of outside columns. 

8. Find outside column loads l^= ; 2 \ ^2 > ^3 "S 

9. For eguo/bays, inside columns get no direct load , For uncgual bays,inside columns 

get direct load P - - S.” V\ where V* = ,k V . 


Note :ln preparing a wind stress diagram, the following items only are givenon the dia^^m: 

1. Beam moment for each floor level. Constant tor same floor. 

2 . Moments in exterior columns. The moments in interior columns are twice os great. 

3 . Direct ovifll loads in exterior Columns,also in interior columns when bays are unegual. 


68 


D.Molifor,C.E, 

Aug. 1926 . 











































































Art. 34. More Accurate Solution for Bottom Stories 
of a Frame~in donjunction with an Approximate Solution 

o7 tlie tippe r Stories > 

Should it be desired to solve the first three 
stories of a frame with greater accuracy than that 
afforded by the approximate design, then the following 
method, illustrated in Fig. 34-A, may be employed. It 
does not depend on assumed points of contraflexure of 
the columns, but does embody the other assumptions as 
to distribution of the total wind shear among the 
several colximns, and that the beam moments are equal 
in the same floor level as in Fig. 33-A. 

The method is the same as given in Problem 
29-H, except that the effect of the upper stories above 
the fourth floor is represented merely by the total wind 
shear from the roof down to the point of contraflexure 
of the fourth story outside column A. This is possible 
because the sum of the moments of all the wind shears 
and beam shears acting on the exterior columns, 
must be zero about any point of contraflexure in the 
column. 


Hence in Fig. 34-A, the total horizontal 
shear H , on the outside column, due to wind from the 
roof down to the fourth story, is the resultant ex¬ 
ternal force acting above the foxirth floor and must 
be considered in producing the moment area. 

The error in assuming H to act at ^114 above 
the fourth floor, is relatively* small and becomes 
negligible for the lower stories. 


69 






•iK 


MORE ACCURATE SOLUTION FOR THREE BOTTOM STORIES OF A FRAME. 

INCONJUNCTION WITH APPROXIMATE SOLUTION OF UPPER STORIES. 




j^4tf».f loor 


"3 

A wz 

lx, 

3'^Story 
c .floor 


Hz ^ 

jx. 

2»’d.siory 

loor 


h, 

k 

1 St. story 

vr 1 St. floor 


> 

LxbaZZ'-. 

18 ' — 22 ' -£ 



►■b 


* D 


X=l-lb. 

I * ^ 









\ 

m* 


m, 

zb'’ 

.b, 


A 


A 


Xa( 


Four boHom storie s of < jtven fram e, 


FIG.34A. ® 


M^areo, 


V 

m^ oreg . 


V 

W^Qrea. 


X,= l 
w oreg. 


The wind loods f W'^ ^ and ihe hor. ih&ar H , are ^hose acVirxc^ on Column A only. 

The ordinates gf-^he M a Orea ore the Contile/er moments of fhe wind loads w,, ond Wj , 
and fhc shear H abouf f-he t st., 2 Qd, and ^jh floor 1 e^c) s, respec+i ve ly, assumin^^ the fro me 

cuf by o verf I cal Sccfion throu^jh fhe points O'f confraflexure of the beams in the l^^span. 

All external Forces act on column A as a principal system lyith moments as foilo*«/s: 

^^3+- >^ 3 ] . __ . 

M, = H[h^+h 3 -*-+ M/ 5 [h^++ w^Ti^ •, H[hj+.^ 4 ] jond ^ 4 . J %** 

The FLASTICITY EQUATIONS Forthe 3 unknown sheors X,,X^ond X^^ore; 

X,ym*dx -r ■*" dx = /m„ m^dx 1 

X,y'nijm|dx + dx + )(.^J’vn^m^dx rr y^M^trij^dx >--£^' 2 .. 

X,y^m j m,dx + *’’z dx +X 3 yf»iJdx riyM.mjdx J 

The integrals in these F< ^S. 2j are evaluated once for all as follows: 

; _ _ /m^d«=-b*(h+^) 

M.Wjd* = ^<(ivi_+Mj; ym^dnsltth.^hj* 

ym,m^dx = h, b*" j j*r\^m.^dx =h,b^; and y^m^m^dx =1 (h, + h^) . 

Column Moments . Bcam Moments . 

2 * 2 ^Floor M, = "bX, 


>•-F<f3. 


= ■btXj+X^+X,) . M,-b(X 3 -»-X 2 ). 

M;=IV1, - To(X3-i.X 2^X,) i Mj=Mj-'bX5 . 
M,=M|- ^(Xj-rXi) M3=ivi3 -bX^. 


3 !!^ Floor b X^ 

4ti! Floor bXj 


-E(p4. 


PROBLEM : Fii^.s, p zo-Bull- 40 ,Univ.of III.^M^ilsond; Money. Totol5hear4^5tory = 6Z70*ar H»£322= 104 ^. 
The wind loads w,,w^ , for Col. A only, are '/6 the total wind load of the respective floors.^ ^ 

I n serf in<] these loads and the dimensions intotheabove M^orea^then the inte^^rat Eqs .3 Can be solved 
to obtain the foliowin<] numerical elasticity E<JS. 2 *. 

3106 X, + 2662 X 2 + Z 662 X 3 = 13,553,210 Solved with f X, = 17^0 Ibs. 


2662 X,+ 5042X2+ 4598X3= 19,26 1,240 
2662 Xif 4S98X^+ 6736X3= 21,590,88 

Hence., M^=7o.c>8S-i|x48oo = 17.28Sft.lbs. 
from ■ M* = 41,9X5-11x4800 ss_ 10 575 ~ » 

EqV4j M, = 4|,925-|U3DI 0 = 6 , 8 lS •* « 


A io'*Slide rule 
qive ; 

M.j^«X2940-H k3o lo = -)o,l 7 oft.lbs 

M2=^^2940-11X 1420- 7,320 •• " 

M^» 7315 -II *1420= - 8 3o S- •• 


X^= 1590 
Xjr 1420 


m.s 19,700 H.lbs. 

J7.SOO" « 
tn^m ) 5,620 x « 


70 























































Art. 35- Comparison of Results due to Several Methods* 
Table 35 A was compiled to illustrate relative" merits of 
results obtainable by the four methods compared. The 
Slope-deflection analysis was taken from Bulletin 80, 

Eng. Experiment Station, Univ. of Illinois; the Ross 
figures are from a paper on "The Design of Tall Building 
Frames to resist Wind" by Ross and Morris in Proc. Am. 

Soc* C. E. May, 1938; and Fleming’s method I was used in 
preference to the less rational Fleming Methods II and III. 

It should be noted that in the Slope-deflection 
solution as well as in the Ross and Morris Method, the 
larger discrepancies, especially in the column moments, 
are due to the abrupt changes in moments of inertia of 
the columns, where the section Just below a splice is 
always of excess area. In all approximate methods, 
where moments of inertia are not given much consideration, 
such effects are obscured, as well they might be. 

A careful excunination of the Table 36 A,will 
show that there is little, if any, choice between the 
tabulated results, hence, from the designer’s stand¬ 
point, they are either equally accurate or equally 
erroneous. 


The approximate method in Art. 33, is con¬ 
sidered rational, and more practical than any of the 
other approximate methods* and being based primarily 
on economic considerations as well as simplicity, may 
well be accepted as standard practice. 




13 

12 

II 

10 

9 

8 

7 

6 

S 

4 

3 

2 

I 


TABLE 35A. COMPARISON OF RESULTS FROM VARIOUS METHODS. 

?0 STORY BUILDING^WILSON ScMANEY, FI G. 5, BU LLETI N 40^ UN IV. OF ILL. 
ABULATION OFMOMENTSFOR LOWER I2ST0R1E5 FOR 30LB.WIND0N 1 FT.WIDTH. 


TOTAL 
Vino LOAD 


W 
3N I FOOT 
VIDTH 
50lfaS.S<).ff 


Moments Col.a 
FT. KIPS 


z 

o 

H 

u 

lu 


tu 

a 


O O 
0:: S 


vfl 

z 


X 

0 

I- 

ri 

o 

s 


Moments col.b. 
FT. KIPS. 


z 

o 

I H 
tu KJ 

“is 

O u. 
IM 

»n o 


I 52 
»/> 

(/) o: 

o ® 

O' S 


M 

Z 

S 

Ul 


o: 

o 

I— 

j 

o 


GlROETR MOMENTS M. 
FT. KIPS. 


Lu 

n lu 

Q- Zl 

o Y- 

_J lu 

<0 Q 


«/) 
I — 


q: 2 


H 

d 

z 

i 

ut 


X 

O 

H 

n 

o 

2 


K w - 
g 28do'^ 
^ lbs. 33 


6.12 


6.10 


6.0Z 


6.12 


3240 


3.4-S 


2.84 


3.23 


3.Z3 


3.18 


3.24 


6.79 

6.26 


6.49 

6.50 


6.S9 


6.48 


&AZ. 


6.61 


6»69 


6.84 


36oON 


3.96 

3.46 


3.58 


3.55 


3.49 


3.60 


7.42 

7.60 


7-21 

7.26 


7-34 


7-20 


7.47 


7-43 


726 


7S6 


i 3960f 


3.87 


3S3 


3.86 

3.86 


3.83 


3.96 


8-50 


7.8S 


798 

8.01 


8.07 


792 


792 


6. o6 


6.07 


8-28 


4320 H 
Splice 


4-38 


4;?.? 


4-20 


4-.22 


4.28 


4.32 


8.68 

8.61 


8.77 

8-73 


8.71 


8.64 


7 98 


8.80 




9-00 


0 4680 N 


3,70 

5.64 


4.S8 


4,6g.. 


4.66 


4.6S 


8.10 

10.62 


9.47 

935 


942 


9.36 


i£4S>. 


9.74 


9.S4 


9.'g.4. 


^5040 
Splice 


480 


5.00 


5.0S 


5.07 


493 


5.(6 


10.17 


10.33 


1007 


lO.OS 


(0.17 


10.32 


11-25 


1146 


li-LP- 


H49 


5430 


S.2S 


6:93 


6-39 

g.53 


6.19 


6.33 


11- 94 

12- 97 


12 61 
12.49 


12-75 


12.66 


13.97 




12.90 


I3.ls 


^5850 4 
spitiee 


7.04 


6.90 


7.02 


6.82 


6.74 


6.8Z 


I3.SZ 

13.50 


13.44 

13.63 


13.59 


13.64 


147.1 


14.14 


14.10 


14->3 


h 6270 it 

L/ •“ 


7.32 

732 


7-32 

752 


7.26 


7.3f 


14.67 


14.46 


14.65 


1443 


I4.6J 


14.62 


IS.S9 


IS-g4. 


14.85 


l§JL 


j) 6690 3 
Splice 


6.36 


7.56 


8.02 


7.80 


7.62 


7.80 


15.6Z 


IS.I7 


15.39 


15.63 


15.60 


16.92 


I7.30 


16.65 


17-30 


I 7140 


9-42 




9-50 


lO.lS 


9.25 


9.50 


18-87 


>9-92 


19-05 

18.30 


19-05 


19-00 


2398 


2S.33 


23-27 


23-63 


f77»0 0 


1 




14-90 


22.67 


1388 


22-00 


13.91 


14-13. 


19 . 80 * 


Z0.9Z 


ZS.68 


21-79 


^7.14 


28.60 


28.30 


■26.:go! 


*22 <0.7^12854= 19 800 Ft.lbS. 


22XO.SX I28S X 2= 2S300Ff.lbi, 


72 



















































Art.36. Beam and Column Wind Connections. Any beam 
subjected to floor loads and bending stresses due to 
lateral wind pressure, must be designed for the maximum 
combination of both. Thus for the dead and live loads 
alone on the basis of 18000 lbs. per sq. in. in bending 
and 12000 lbs. per sq. in. in shear, and for 24000 lbs. 
per sq. in. in bending and 15000 lbs. per sq. in. in 
shear for the combined wind and other load effects, 
according to the specifications of the Am. Inst, of 
Steel Constr. 

Fig. 36 A illustrates this beam loading and the 
resulting moments and end shears governing the design of 
a beam and its connections with adjacent columns. 



2 


Fig.36/\. 

The maximum center moment due to floor load is 

for a simple beam without continuity, and the 
wind moment tot the point C is zero. 

The maximum wind moment occurs at the ends of the 
beam and is ± Vi/z , while the floor load moments at these 
ends aiTS zero • 

The maximum end shear is R ^ -f* V. 

Hence the beam must be designed for the center 
moment Me on the basis of 18000 lbs. per sq. in. unit 
stress in bending, and for the end moment on the basis 
of 24000 lbs. per sq. in. unit stress in bending. The 
beam must,therefore, satisfy both moment requirements at 
center and ends, and since the end moment must be developed 
by a suitable end connection with the column, this latter 
condition, involving the strength of the end connection, 
will usually govern the size of the beam. 

73 

























This emphasizes the Importance of the end 
connections and unless these are so designed as to develop 
the full capacity of the beam in bending, then a beam of 
larger size must be selected for which a proper end 
connection becomes possible, and the beam itself will 
then be of excess size and be wasteful* 

Also, since the end moment, due entirely to wind, 
usually governs the size of the beam and its end connections 
therefore, it would be a mistake to increase the negative 
end wind moment by figuring continuity in the beam for floor 
load* In this case the combined positive end moment would 
be diminished, while the negative end moment would be 
increased* 


For beam and Column Connections with knee 
braces the moment conditions are the same as shown in 
^ig* ^6 A, while the stresses in the braced end are as 
Illustrated in Fig* 36 B* 



Jot one braoe as at A; 

Shear at column face = ± taken 

by beam web connection. ^ ° ^ 

= hence T= Vf - tension or compression. 

Stress in brace =: ± TS€c.4S® = Vf sec.4-S* s= ± . 

•zh 2C 

M„= V{i-W)-t and 

m '2. ' z ' ^ ^ 'a 

The points A ond A* are the centers of rivet i^jroups 

at the two Connections with column A. 


74 































The bottom flange of the beam must be connected 
to the column at A for a tension T, and both end connections 
of the brace must be designed for a tension T* and a shear 
T giving a direct stress in the brace of ± 1.4.1 t . 


For two braces as at B; 

Shear at Col.face = ± J token by beam ^eh connection. 

M eV^=:Th\ hence T = stens ion or Compression • 

Stress in brace = ± Tsec.4-S'*= Yf, 5ec.4S*=i ±V6 . 
iVl^s; V(1-dj-h . 

The points B and are the centers of rivet 
groups at the two connections. The stress in the braces 
is again ± 1 . 4-1 T . 


Art.37r Details for Beam and Column Wind Connections. In 
designing tlie beams and their connections, the procedure 
should be as follows: First design the beam for its floor 
load on the basis of a simple span, and then select an end 
connection from Table 37-A, sufficient to develop the 
required end wind moment for this beam. 

In case the actual wind moment is greater than 
afforded by any end connection given in the table for a 
beam of this size, then the beam first selected must be 
increased to such size as the wind moment may require, using 
the strongest detail given for type T connections. 

The smallest size beams, satisfying the center and 
end wind moment requirements, will result in the most 
economic design of the wind frame. 

Table 37-A serves a very useful purpose and, 
except in extraordinary oases, will cover all standard 
connections in tall buildixigs. 


75 





DETAILS OF BEAM COLUMN WIND CONNECTIONS. 



TYPE L 

CONNECTIONS 

TYPE T CONNECTIONS. 


BEAM 

2L?6x4k^-I2'' 

2L^6x4-^%-12 
7/g, Rivets 

2L^8k4k^-I 2' 
l" RIVETS. 

BEAM 

^-271-101-lbs. 
l2"lonq-privets 

^-301-115-lbs. 
l 4 "lonq- Trivets 


I2*CB2^ 

3 2,00 O 

43,00 o 

56,0 00 

121 32* 

72,0 0 0 

_ 


14 •< 30 

31,0 0 o 

50, 0 O 0 

G5,0 0 0 

141 33 

95,0 0 0 

_ 


15 "373 

A 0,0 o o 

53, 0 O O 

70,000 

IS I 42.9 

108,0 OO 

1 18,000 


16 .. 3S 

42,0 O 0 

57, 0 0 O 

15,0 0 0 

161 38 

n 5,0 o o 

118,0 0 0 


00 

4g,0 0 O 

65,000 

84, o 0 o 

I8I47 

1 30,0 O O 

169,000 


20 BI56 

53,0 0 0 

72,000 

93,0 0 0 

2or 56 

143, OO 0 

188,0 0 0 • 


21 CB58 

56,0 0 O 

75,0 0 0 

98,000 

211 58 

151,00 O 

197,000 


22 6158 

58,0 O O 

79,0 00 

10 2,0 0 O 

Z2l 5 8 

158,00 0 

207,000 


24CB70 

64,0 OO 

86,000 

1 1 2,000 

241 10 

172,00 0 

2 26,0 0 0 


27 ■' 91 

72,000 

97,0 0 0 

126,000 

271 91 

194,0 0 O 

254,0 00 


30 .. ns 

80,0 0 0 

\ 01,0 0 0 

140,000 

301 US 

216,00 0 

282,0 00 



Unif s+ress in bendin^j, 
Stress in sin()ie shear 


24000,lbs. per SCJ. In. for wind moments, 
or+ension for !-^ rivet =7,9 5 0 lbs. forivind- 
" •' •' l-%nVet =10,80 0 lbs. •• 

" *' " I - Tri/et =I4. loolbs. 


IE" 


O O O O 






-- 

o o o o 


*_ Column 




o o o o 

e:^r-\ 


1^0 



o o 

i/yhf 

o o 

zSd 

o'o o o 

«— Couu MN 



TYPE L CONNECTION. 


TYPE T CONNECTION. 


% 


S.MoIitor 
















































































Since the columns are usually iz' sind 14” H columns 
from four to eight rivets can be utilized in the connection 
of each flange for type L connections, giving the following 
tension and shear valuesfjfor a group of 4 rivets on the basis 
of 24,000 lbs* per sq. inch, 

4 - 3 / 4 ” rivets <3 7,950 lbs. 31,800 lbs. 

4 - 7/8” " O 10,800 ” 43,200 ” 

4-1” " O 14,100 ” 56,400 ” 

Double these values are used for groups of 8 rivets 
in type T connections. 

The moment of resistance of the end connection for a 
beam of any depth d is represented by the product Pd in 
foot lbs. as given in Table 37-A. The maximum value which Pd 
can have for any particular beam is M = fS= S , whereS 

is the section modulus of the beam in inch units. It should 
be noted that the light weight sections are usually strong 
enough to develop a resisting moment eq.ual to or greater than 
that of the best possible end connection, hence only light 
sections should be employed except when the floor load reciuires 
a heavier section. 


There is one other point to be covered before con¬ 
cluding this subject, and that is the required thickness of 
metal in the connecting angles or tees, to develop the assumed 
rivet values P. 

For type L Connections. 


C__ 



Fi|. 2>7A. 
by the unit etresB 24 


Let 

C=length of connecting 
angle• 


thickness of angle metal 

P= tension or shear value 
of a groupe of 4 rivets 

Assuming a point of 
contrariexure at c, 
then the bending moment 
in the angle leg will 


be ^( 2 -t) 


ind 


ooo 


this must be equal to 
the resisting moment 

multiplied 
Henoe the following equation 


^ X 2+000 , or 


P- ^=+000 ft*" 


which solved 
For ?= iz" . 


for ir gives 
P= 31,800 lbs., 
P S 4 3 zoo •• 
56,40 0 M 


oW 

P ' 

p 

f4oooP \ 

I60006. 

/ \6ooof 

0.64, 'inch 

for 

4 ->/+ riVcts 

0.70 »* 

for 

4- 7/s rivets 

t 0 . 7 S •* 

for 

4- 1 rivets. 


-- 37A. 


77 



















For type T connections 



the heavy fillet is deducted 
from the span between rivets 
to obtain an effective span 
S = 5 -1.8 = 3. 2, inches . 

The center moment of this span 
with restrained ends and due 
to a load P will be 

^ 54200 in.lbs 

The section modulus of the 
flange metal is i z x o.~S3S _ f 
c 

giving a unit stress in bend¬ 
ing of 

fsJliLst 5±2£2. = 24300/bs. per M.m. 

S ‘-404 ^ ^ 


The oase with a half 3o 11 is lbs, using 1” rivets or 
Pa M 2 . 8 po lbs. gives a unit stress f= 23900 lbs. per s<^.in. 


Art.36. Wind Pressure. The municipal buildir^ codes 
stipulate the wind pressure to be used in designing tall 
building frames, chimneys, sign boards, etc,, but in the 
fiibsence of any such requirements, the designer may have 
to decide the question on the basis of Weather Bureau 
observations. 

In deducing wind pressures from velocity obser¬ 
vations many difficulties are encountered even if the 
velocity were known with accuracy. Wind normally consists 
of a series of gusts, the speed and direction of which vary 
within wide limits. 

The U. S. Weather Bureau furnishes wind velocities 
for many stations in the country and covering a long period 
of time. Since maximum velocities occur only at intervals of 
many years it is essential to select the extreme maximum for 
a ten to twenty year record for any locality. 

The Weather Bureau velocities are somewhat too high 
according to experiments conducted at the U. S. Bureau of 
Standards, published in Scientific Paper No. 523 by Dryden 
and Hill. The true velocity was found to be 76^ of the 
Weather Bureau velocity. 

Records show velocities ranging from 60 to 120 
miles per hour for various stations in the U. S., but it 
would not seem prudent to figure less than 60 miles even 
though weather reports showed a lesser maximum. 

Except in some of the storm swept areas on the 
Atlantic Coast and the Gulf of Mexico, 100 mile wind is quite 
conservative and 87 mile wind is a fair maximum for all 
inland points. 


/ 




78 























The generally accepted formula for converting 
velocity into pressure is that of E. Eiffel, as translated 
by Hunsaker, U. S. Navy, 1913, 

vvr=iC3. 0033 V^lbs.pei^ ft-- --- — 3g/^. 

whdre V is wind velocity in mil^s per* hour. 

Whether the wind pressure requirements should be 
varied to suit the ratio of height to width of a building 
is questionable, because there are more important factors 
to be considered besides this ratio* 

For strength alone, the specified unit stresses 
govern every design. However, rigidity is of even greater 
importance than strength, and this property depends on the 
dimensions of the frame and the method of carrying the wind 
loads* 


On the assumption that the bare frame should 
sustain the entire wind load without exceeding certain 
allowable unit stresses, then the most rigid frame will 
be one of rather wide base compared with its height, and 
the wind shears are carried by diagonal members in direct 
stress* This, however, cannot be realized except in very 
rare cases, because diagonal members must be concealed in 
hollow walls which latter are very objectionable from the 
architectural standpoint* 

^ In the usual type of tier building construction, in 
whichlrrame consists of beams and columns without diagonals, 
wherein all wind shears must be resisted by bending, the 
inherent rigidity of the frame is due to the number of 
columns of given story heights, and not to the ratio of 
height to width of frame alone* Of course a wide building 
usually contains more columns than does a narrow building, 
nevertheless, it is the number of columns and the rigidity 
of the beam and column connections which determine the 
rigidity of the frame as a whole* 

The very considerable contribution to rigidity 
attributable to the architectural fill materials will usually 
afford ample security against excessive deflections as men¬ 
tioned in Art* 31* 

The practice of introducing an indeterminate 
factor, by assuming the upper six or eight stories of a 
tall building as contributing no wind load beaause of the 
architectural fill material, is not considered wise* 


79 



Chapter 8, Apprexlaate Method of 
Desi g ning Continuous Concrete Beams . 

David A. Molitor, C. E. 

Art. 39. Description of Method. The method here advocated 
for the design of continuous concrete l>eams is applicahle 
only in building constructions where beams are of T shape 
and are cast monolithically with the columns. Such beams 
involve a large element of uncertainty which cannot be 
appraised by any practicable theoretical analysis. 

The method consists of design ingf continuous beams 
like cantilever systems, in which the more or less arbitrarily 
located hinged points are treated as points of contraflexure. 
To make some allowance for shifting positions of the live 
load, the design moments are increased 20^o over the actual 
theoretical moments of the cantilever system. The method 
has been used in Detroit, Michigan, for over ten years in 
the design of reinforced factory and office buildings and 
has proven eminently satisfactory. 

The formulas for cantilever beams will first be 
given, together with a table of moment coefficients for end 
and interior spans. An illustrative problem will then be 
presented followed by a discussion of the method. 

For any case of uniform loading the sum of the end 
and center moments for an interior span will equal iwBVs# 
and adding 205 ^ to each will give the sum of the negative end 
and positive center racmients of 1.2 x o. 125 

For end spans this sum is not a constant but increases as the 
point of contraflexure moves farther out from the restrained 
end. 


The formulas are given in Table 39-A, and may be 
written out directly from Fig. 39-A. The formulas in the 
left half of the table represent the theoretical values, and 
those in the right half are wrking values increased by 20^ 
to allow for variations in the points of contraflexure as 
affected by different cases of loading. 

It should be noted that the locus pf the point of 
contraflexure for an interior span is a parabola representing 
the constant sum moment o.i5u>£* , wftiile for an end span the 
locus of the point C is a straight line, such that for every 
point C , there will be a separate parabola passing through it 
to represent the end and center momants. 

Fig. 39-A is drawn for one particular set of 
moments, with the moment over a support representing the end 
moments of any two adjacent spans. The same feature may be 
preserved even whan the spans are of unequal lengths or when 
the loads on the spans are widely different.With this arrange¬ 
ment, no moment is transmitted to the column between any two 
spans. 


80 








CONCRHE CANTILEVER BEAMS ,UNt FORMLY VOADED . DETROIT CODE. 


END SPAN INTERIOR SPAN . 



rORMULAS . 


MOMTNTS FOR ALL SPANS LOADED 

INTERIOR SPANS . 

Me = V^c(l-2c)^- 

M^-f Mg = 5^3- 51 constant. 

2 f:4 wu 2 2 

END SPANS. 

Me = i|^(l-c.)+ 

M- = wcn-c) 

21 ^ 

ryi^ + Mg = .^(l+ c)^= variable 
C = 

0.5 tul 


MOMENTS INCREASED 20%> . 

interior spans . 

o.isw;( 1 - 2c)^ 

Mg = o.6toc(l- c) 

+ IVlp =1 0.|5 lA) constant. 


2 oiSiAJ 2 2 Vo. 


v^' 

END SPANS. 


Me 

ISve 


o.is(l- ^)*’yvl 
Mg = o.s wcl 

M = QJ^e^l-c) = 0.3uoc(l-c) 

M-rMg = 0.iswr(l^- c^) 

o.jsiAJ (l+c)^= variable. 


c= 

0.6 u;l 


See Trans. Ann-Soc. C*.E , V 0 I. 91 , p. 174 ,'Paper entitle^i *Diitribu+ian oi Reinforcina Steel 
in Concrete Beams and Slabs’; by Boyd S. Myers, av.d Discussion on p-ziz,by!D.Molitor. 

B. Molitor, C.E 
Oct.2,1926. 


81 
































TABLE 3BA. 

END AND CENTER MOMENTS FOR CONTINUOUS BEAMS. 


END SPANS 

END MOMENT 

CENTERMOMENT 

Point 

Contra. 

flexure 

Fraction 

Decimal 

Fraction 

Decimal 

WL^ 

’/ 19.2 

WL» 

0.’Q522 

WL? 

'/8 

WL' 

0.1250 

L 

0.087 

»/l8.5 

0.0 540 

78.05 

0.1242 

0.090 

•/ 17.5 

0.0S70 

7 a. 15 

0.1229 

0.095 

‘/I 6.7 

0.0600 

/8.25 

0.12IS 

0.100 

‘/l5.9 

0.O630 

78.33 

0.1202 

0.105 

>/l5.l 

0.06G0 

78.43 

0.1188 

o.no 

‘/ 14.5 

0.0690 

78.52 

0.1175 

0.115 

!/|3.9 

0.0720 

78.62 

0.1162 

0.120 

’/l3.3 

0.0750 

78.73 

0.1148 

0.125 

>/l2.8 

0.0780 

78.82 

0.1135 

0.130 

Vn.e 

0.0840 

79.04 

0.1109 

0.140 

'/ii.i 

0.0900 

'/9.24 

0.1084 

0.150 


0.0960 

79.38 

Q.1058 

0.160 

V 9 .S 0 

0.10 20 

79.69 

0.1033 

0.170 

/9.25 

0.1080 

79.92 

0.1009 

0.180 

Vs.7 8 

0.1 140 

7lO.)S 

0.0985 

0.190 

'/8.33 

0.I2OO 

7io.4 

0.0960 

0.200 

V6.G7 

0. J500 

7 n .8 

0.O84S 

0.25 

•/S.55 

0.180 

7i3.6 

0.0735 

0.30 

'/4.76 

0.210 

yi5.7 

0.0635 

0.35 

VA.\& 

0.240 

7i8.S 

0.0 540 

0.40 

'f^-70 

0.270 

722.0 

0.0454 

0.45 

V 3.33 

0.300 

725.7 

0.0375 

0.50 

’/2.78 

0.360 

741.6 

0.0 240 

0.60 

‘/2,38 

0.420 

774 

0.0/35 

0.70 

V 2 06 

0.480 

Vl67 

0.0060 

0.80 

Vl-85 

0.540 

7666 

0.00 1 s 

0.90 

'/I.67 

O.60O 

7oc 

0.0000 

).oo 


INTERIOR SPANS. 


END MOMENT 

CENTERMOMENT 

POINT 

Fraction 

Decimal 

Fraction 

Decimal 

contra. 

flexure. 

WL* 

'/40 

YYL^ 

0.0250 

WL’- 

Va 

WL‘ 

0.1250 

L 

0.04S 

'/35.I 

0.0285 

78.25 

0.12 1 5 

0.050 

732.0 

0.031 2 

78.40 

0.1188 

0.0S5 

'/i9.4 

0.0340 

78.63 

0.M6O 

0.060 

727.4 

0.0365 

78.82 

0.1135 

0065 

725.6 

0.0391 

79.03 

0.1 109 

0.070 

724.1 

O.04I6 

79.23 

0.1084 

0.075 

722.7 

0.0440 

79.45 

0.1060 

0.080 

721.5 

0.0467 

V9.68 

0.1033 

0.085 

720.3 

0.6492 

79.93 

0.1008 

0.090 

7i9.4 

0.0516 

7i0.2 

0.0984 

0.095 

V18.5 

0.0540 

710-4 

0.0960 

0.100 

7»7.7 

0.0564 

Vi 0.7 

0.0936 

o.ios 

Vito 

0.0587 

7io.9 

0.0913 

0.11 0 

716.4 

0.0610 

711.2 

0.0890 

0.1 IS 

yi5.7 

0.0634 

Vii.s 

0.0866 

0.120 

7l 5.3 

0.065S 

711.8 

0.0 845 

0.125 

'/l4.7 

0.0 678 

’/12.1 

0.0822 

0.130 

'/l4.3 

0.0700 

7l 2.5 

0.0800 

0.135 

7I3.8 

0.0723 

712.9 

0.0777 

0.140 

7|3.4 

0.0744 

Vi 3.2 

0.0756 

0.145 

7|3.| 

0.0765 

Vi 3.6 

0.0735 

0.150 

/12.7 

0.0786 

V14.0 

0.0714 

0.155 

712.4 

0.0805 

7i4.4 

0.0695 

0.160 

'/I2.I 

0.0828 

714.9 

0.0672 

0.165 

7n.8 

0.0846 

7 IS .3 

0.06 54 

0170 

7ii.6 

0.0866 

715.8 

0.0634 

0.175 

7m.4 

0.08 75 

VI6.0 

0.062S 

0.177 


li. Mollto r. 


Tbr Interi or Spans ihe center mom. ^hall be not le5s than WI.VI6, and the end mom Shall be 
not less than WLV40. 

for End St>Qns the end moment shall be not less thon wC'/zo.and may be anytfn'nq 
depending on odjacent Spans. ^ ^ 

iSa^td A-Molitor.C. E. 

Oct. 2,1916 


82 
































































































'lX.'bOO 


Table 39-A is self explanatory and will suffice for 
all spans of variable lengths and loadings. 

In the following example no attempt is made to delve 
into the theory of reinforced concrete design, but merely to 
illustrate the method of arriving at a suitable assignment 
of end and center moments in a series of continuous beams. 

The standard nomenclature used will be found in any hand book 
on reinforced concrete. 


Art. 40. Example 40«A. Three spans have uniform loads of 
2£00 lbs. per lin. ft. and concentrated loads as shown in 
Figi 40>Ai to design the beams on the basis of the following 
unit stresses: f^= 750 lbs persqr inv fs-20,000 Ibs.persa.'m., n=i 5 , 

= 118.8, and the moment of resistance s m.p ^ 
ft. lbs. for concrete sections. The allowable unit'^hear is 
not to exceed 120 lbs. for a section with bent bars and 
stirrups. The beam has a 4 inch solid al4b flush with the 
top and will act as a T beam. The end shears for each span 



First choose a beam section adequate for the 
maximum end shear of 32jOOO lbs. which occurs in the center 
span. For this shear make the beam 12" x 30" giving a unit 
shear =n2 ibs. Then M.R. = li^bcl‘s-93,ioo f+.lb6. 

for'the stem witnout using any compression steel. This 
moment requires a tension steel area Ag= 2 ~^ -? ? 73>7 in. 


for which use 3 - 1 in. round bars. 


This section ?dll be accepted for all three spans, 
making eill end moments equal to 9^100 ft. lbs. and the 
corresponding center moment for each span will now be found 
using the moment coefficients given in Table 39-A. A T beam 
section of 33 in. width of T will be ample for the max.M^= 204,000 

''em Span 1. Max.M=220o x sooo = III, 800 ft. lbs. 

For this moment make tos><.111,800= 894,000 ft.lbs* 

For an end moment Me = =93,ioo, fi*nclx=9.ss 

From the table for end spans, for an end moment with x=9.65r 
find the corresponding center moment with x*9.75 and c/e- 6.172 

Hence M = g94.ooo ^ 92 goo ft.ibs., Qwd 0 = 0 . 172 ?= 3, 1 -Pt. 

^ 9.75 

The required steel area ^ 5 = s<{.in » 3 —. 

1" round bars. Make one straight (full length); one bent 
for shear; and one bent for double continuity over col\imn2^ 


S3 
































expecting to get two more 1” round continuity bars from 
span 2, 


Int. Span 2 . Max,Mr iliJ = 247,500ft. lbs. and wf = i, 980 ,ooo ft.Ifa*. 
^OT an end moment Me = 93,100 ft.ibs., f/nd x= 21 .3 

and the center moment 


M 


= 204,000 ft.lbs.and C = 0 . 085 x 3 o= 2 .SSff, 

x=9.3 


The required steel area = ^o4gog s5r.o5q.in.Use 2-1 in. square 
bars (full length); two 1 in.^ound bars bent for shear; and 
2-1 in. round bars bent for double continuity at each end 
of span 2. 


Figuring a 4 inch T, 33 in. wide^the moment of 
resistance of the T beam = 204000 ft. lbs. as required. 

End. Span 3 . Max.M=28,40oxi2'.9-22ooxJL^= »84oooft.lbs. 

For this moment make sx i84,ooo = j,472,ooo fi.ib*. 

For an end moment Me= ^* 93 100, find x= 15.95. 

From the table for end spans, for an end moment withx = 15.95 
find the corresponding center moment with x=:8.33. 

Hence M<.=177 ,000 ff.lbs., and cso.iosx24= 2.5 ft. 

The required steel area2-1 in. square 
bars and 3-1 in. round bars. Make 2 - l'^ square bars 
straight; 2-1" round bars bent for shear; and 1-1" round 
bar bent for double continuity over column 3. 

The bars bent for double continuity are bent up 
at the point of contraflexure at 460 and extend into the 
adjacent spans 1 ft. beyond the respective points of contra- 
flexure nearest the coliimn support. It will be seen that 
the foregoing arrangement of bent bars furnishes three 1 in. 
round bars over the top of each support as required for 
the end moments of 92^100 ft. lbs. The bars bent for shear 
are bent up at 450 next to the columns and at the outer 
ends and are hooked into the columns and at the outer ends. 

No extra bars are thus required for the negative 
end moments as the necessary steel was obtained by proper 
bending and careful selection of bar sizes to make the 
steel areas correspond with, the computed areas at all beam 
sections. Also, the end moments on each side of each column 
being equal, there will be no moment transmitted to the 
columns except a very nominal moment which may result from 
partial live loads on any of the spans. 

In the foregoing example, say 2.0 sq. in. of 
compression steel might have been utilized by extending 
the bottom straight bars into the columns and adjacent 
spans, making the negative end moments 129,000 ft. lbs. 
instead of only 93,100 ft. lbs. as furnished by the stem only. 
This would have resulted in somewhat smaller center moments 
and a slight economy in steel. 


84 







If, for any reason, the end moments of an 
interior span are not made equal, a condition which is 
governed by the lengths and loadings of the adjacent spans, 
then the center moment is based on the mean value of the 
two unequal end moments. Similarly, if a certain end 
momant is assigned at Col. 1 or Col. 4, to be transmitted 
from the end spans to the end columns, then such end spans 
are designed for center moments based on the average end mo¬ 
ments of these spans, the same as for interior spans with 
end moments. 

The shear reinforcement still to be provided for 
in the form of stirrups is computed on the assumption that 
the concrete is permitted to take 40 lbs. of shear per sq. 
in. of stem area, and the bent shear bars are good for the 
added shear increment between the point of the first bend 
and the end of the span. 

Thus for span S, the gross end shear is 33000 lbs. 
the stem area takes 4 o bjd = m,7So lbs. and the bent bars 

will carry 3 x 3300 = 6,600 lbs. leaving a net shear Vri4,G.4o 
lbs. for which stirrups must be provided. 

ma 

In present. This example, sufficient descriptive 
matter was deemed necessary to illustrate the use of the 
table. In practice, however, the complete numerical work 
can be reduced to about six lines for the 3 span problem. 

Art. 4-1 . Justification of the Method. On strictly theoretical 
lines there is some objection to this method of designir^ 
reinforced concrete beams with continuity, on the presumption 
that beams so designed must function strictly,in accord with 
the theory of elasticity. Yet no tests made to date prove 
that reinforced concrete beams, designed otherwise than by 
the theorem of three moments, have proven unsatisfactory in 
service, and such service covers periods of ten to fifteen 
years on many large and important buildings. 

No objection,other than theoretical, has ever been 
raised against this method of design, and perhaps too much 
weight has been given to matters which are beyond the 
possibility of correct theoretical analysis. If such beams 
have proven satisfactory under normal working conditions in 
buildings,that fact should be accepted as a test result, 
irrespective of what would happen under tests carried to 
ultimate failure. 

In practically all tests made on continuous concrete 
beams designed according to the elastic theory, the steel 
over the supports slipped to some extent and was never fully 
stressed even when forced into action by the positive moment, 
indicating that some extra steel should be supplied at the 
center of the span where slipping cannot occur to make up for 
the loss at the ends. 

In reinforced concrete buildings, the beams are 
T shaped and monolithic with floor slabs and' columns, with 
reinforcing steel so interlocked as to prevent unsightly 

S5 




cracks. The superimposed loads are of extremely variable 
character, uniform dead and live loads and various concentrated 
loads, all-,of which may attain the maximum values assumed in 
the design, or they may be greatly reduced when there is no 
live load. It would be futile to attempt a rigid analysis 
of such a'beam, or system of beams, in accordance with the 
theory of elasticity and hence the need for a simple 
approximate method. 

A T beam subjected to negative and mcments is good 
for no more than the resisting moment of the stem at the ends, 
while the center portion carrying the positive moment, may 
have double this capacity in compressive area of the slab 
which forms the T. Therefore, it is wasteful to bolster up 
the end section by haunching or using compression steel, 
which must be done if the end section is to carry more moment 
than the center section. 

By ignoring the theoretical moment distribution 
indicated for continuity, and treating the chain of spans 
as a series of cantilever systems, in which the hinged points 
represent arbitrarily assigned points of contraflexure, the 
above mentioned difficulties are obviated. 

How it may be contended that, because the points of 
contraflexure are treated as imaginary hinges of a cantilever 
system, which hinges have not actually been provided in the 
construction, that it is not certain that the structure will 
behave in accordance with these assumptions. 

Anyone of the various assumptions that may be made 
as to moment distribution in this analysis, represents 
possible conditions for some cantilever system, and if the 
material (both steel and concrete) at any section of a beam 
is ample to care for the moment assigned to it, each such 
section is capable of contributing its share of total internal 
work to be performed. 

The possible shift of a point of contraflexure can 
be reduced to a minimum limit by bending up part of the 
positive steel so as to have the first bent bar intersect 
the beam axis in the point of contraflexure and extend out 
into the adjacent span for a distanced/zbeyond the adjacent 
point of contraflexure! This is called bending for double 
continuity (D. C.) Other bars bent merely for shear, may 
also act for continuity if allowed to extend beyond the 
supports into adjacent spans. This is called bending for shear 
and continuity (S. & C.) 

Since the loads applied to a beam will first stress 
the central portion containing the positive steel, before the 
load effect can reach the supports, it follows that the 
positive steel must work to capacity, delivering to the 
cantilever ends only that portion of the load in excess of 
its capacity and for which the negative steel makes ample 
provision. 


86 


On the other hand, enough negative steel may be 
provided to carry the entire load on cantilever arms, thus 
unloading the central portion vheve little or no positive 
steel would then be required. However, this alternative is 
not utilized except In special cases of short interior spans 
or short end spans, when the end moments from the longer 
*4Jaeent spans, are proportionately large. 

?ftiil« continuous beaims, designed according to the 
theory of elasticity, develop moments approximating the 
designed values within the range of proportionality, that 
does not prove (as many assume it does) that for beams 
designed otherwise the results under tests would disagree 
materially with the assumptions on which the design was 
based. In fact, the few tests reported by Prof. Moersch 
distinctly support this latter view. Although the behavior 
of the beam within the range of proportionality now becomes 
somewhat different, the ultimate behavior at rupture again 
agrees with the variant assumptions. 

After the elastic limit is exceeded, the elastic 
theory no longer governs, and entirely new relations are set 
up. For this there is abundant evidence which has not 
received proper consideration by theorists. 

Ordinary steel elongates about 26)^ of its length 
at the ultimate stress, while the elongation at the elastic 
limit amounts to less than 0.1^. If then some of the steel 
fdiich is stressed beyond the elastic limit is permanently 
stretched 1^ due to overloads which have stressed steel in 
other parts of the structure, involving redundancy merely to 
the elastic limit, then a readjustment of stress will take 
place resulting in all steel stressed equally. In the case 
of a reinforced concrete beam all this can take place with¬ 
out showing visible cracks over the supports. When cracks 
do appear it is due to much greater overloads, approaching 
ultimate stress, or to other causes attributable to methods 
of construction. 

Advantage can be taken of the ductility of steel 
only when dealing with structures involving redundancy, 
and since continuous beams belong to this class, it is 
proper to give consideration to this steel characteristic. 

Take, for instance, a redundant tension member 
in a steel frame or truss, and suppose that its length is 
shorter than intended,, and the member is forced into 
place and its ends are then permanently connected. When 
the frame receives its maximum load, this particular member 
will be overstressed before the^ other members attain their 
calculated stresses. The stress in such a redundant member 
may readily exceed the elastic limit, and if it does, this 
member simply stretches a small amount (permanently) and 
thus adjusts itself to the defects in construction without 
producing any permanent damage. Thereafter, for repeated 
applications of the maximum load, the member will function 
normally without ever again being stressed beyond its yield 

point. 


$7 


This is a well known phenomenon, hut, strangely 
enough, it has received little consideration in practice. 

A better iinderstending of this subject would relieve most 
doubts now entertained by many engineers with regard to 
statically indeterminate structures. 

The following statement by F.v. Emperger in Beton 
u. Eisen, June 20, 1930, Yol.EQ, No,12, p.216, confirms the 
above conclusions, "The moment curve can be exactly de¬ 
termined only for simply supported beans, \rnere restraint 
exists the moment variation may be studied by considering 
possible limiting conditions of restraint. The usual 
assumption of point supports, satisfactory in bridge work, 
does not seen reliable for buildings v/here the columns act 
with the beams". 

"Building Departments of Leipzig and Hamburg have 
now (1929) approved procedures along new lines. Safety 
against failure should be the governing factor in beaiix 
design. The siim of the center and support moments rather 
than moment at any point is to be considered. Attainment 
of the elastic limit at any point leads to a readjustment of 
the moment curve, causing higher moments at other portions of 
the beam. In concrete, flexibility of design possible in 
reinforcement and haunching, permits original moment dis¬ 
tribution between center and sux)ports to exist until failure 
occurs," 

"In tests by the Austrian Reinforced Concrete 
Committee, (Heft 4, 131^) a fully restrained concrete beam 
v/as reinforced for gDf at supports and at the center. 

At failure, the 24 moments had ^ adjusted them¬ 
selves to fit this steel distribution." 

"At some future date it is hoped to continue tests 
of duplicates of single span, restrained beams already 
studied, when such beams are parts of rigid frames." 

The procedure advocated In this chapter, for the 
design of reinforced concrete beams, may be restated as 
follows: 

For continuous beams treated as cantilever systems , 
the end and center moments may be proportioned as follows: 

For Intermediate suans . the end moment shall never 
be les^ than and the center moment shall never be less 
than whlle^ the sum of the mnan end and center moments 
shall be not less than O.lSwl* . 

end spans , the end moment shall be not less 
than 5^ and may have any value greater than this, depending 
on 13i#^end moment of the adjacent span. The stun of the end 
and center moments for an end span shall be not less than 
0,15 w(1tI^ c)^, where c s . the distance from the 

fixed end to the point of coaferltflexure of the end span. 


88 







CHAPTER 9 - RETAINING WALLS 


Art> 42 Earth pressure on the back of a wall. It is 
not proposed to enter upon a theoretical discussion of this 
subject but merely to present, in brief form, the most 
usable features of the many theories evolved in the past 
by Coulomb, Ponoelet, Soheffler, Mohr, Rankine, Winkler, 
von Ott, Weyronch, Mueller-Breslau and others. Of these may 
be mentioned Coulomb's theory of the prism of maximum pressure 
and the theory of conjugate stresses by Rankine, as the two 
outstanding ones, the combined features of^.^hich will be 
employed in the following outline, together^Poncelets graphic 
solution. 

When an earth embankment is artificially produced by 
dumping the material loosely from dump carts or flat cars, 
the material assumes a natural surface which is usually not 
a plane, nor does the averaged plane of slope make a constant 
angle ^ with the horisontal as is commonly assumed. Actually 
the angle varies, becoming smaller with increased height of 
embankment. For certain clay materials the angle varies 
from 450 for heights up to 6 ft. and reduces to about 18® for 
a height of 66 ft. See paper by the author* "The Present 
Status of Engineering Knowledge Respecting Masonry Construc¬ 
tion." For clean sand the conditions are most favorable and 
the angle of repose § varies only slightly for a wide range 
of height. Other materials manifest a behavior intermediate 
between sand and clay. Laboratory experiments on dry material 
have no value owing to miniature proportions and doubtful 
degree of moisture which a given material should possess when 
exposed to the elements for a long period of time. 

When an embankment is filled to a steeper slope than 
is naturally possible for the material to maintain permanently 
by internal friction, then a slide will occur and the sur¬ 
face slope will adjust itself to a flatter angle. While this 
readjustment is taking place the surface material will move 
on a surface of rupture formed in the interior of the fill. 

The material beneath this surface of rupture will remain 
undisturbed, while the original surface slope will gradually 
assume a position of rest approximating the angle of repose. 
The surface of rupture is never a plane, being approximately 
hyperbolic for clay and approaching a plane for clean dry 
sand. Howerer, a plane surface of rupture is usually assumed 
in dealing with the subject along theoretical lines. 

With these introductory remarks, based on many years of 
careful observation, it should be apparent that any attempt 
at a purely theoretical analysis of this problem is quite 
futile and the most that may be accomplished is to arrive 


*Journal Assn. Eng. Soc..l900, p. 31 containing a new Theory 
of Earth Pressure for Clay. 


89 





at a rational approximate method deToid of unneceeeary 
mathematical manipulations. The natural surface of repose 
and the surface of rupture will be assumed as plane surfaces 
in the following presentation. 

Art. 43 Earth Pressu re on walls without surcharge. 

Pig. represents the case of active"^'earth pressure (a), 

and that of passive earth resistance (b), considering these 
forces without regard to the stability requirements to be 
fulfilled by the well structure itself. 



Fi^S. 43A. 


The active earth pressure E is the resultant pressure which 
the wall Ab, Pig. a, must exert against the wedge ABL to pre¬ 
vent eliding on the plane of rupture BL, when the wedge is 
acted upon by its own weight P. The resultant pressure P, 
making the angle § with the normal to the surface of rupture, 
represents the force necessary to resist the wedge pressure 
from below. The three forces P, E and P must be in equili¬ 
brium as indicated by the triangular force polygon. The 
maximum value of E occurs when the angle + which 

signifies that the surface of rupture bisects the angle 

which is the angle between the surface of repose, and the surface 
AB or back of the wall. In this case the pressure wedge 
ABL must beoome active be fere the wall is celled upon to 
offer the maximum resistance E. 


90 























According to Coulomb the horizontal componcrrtof the active 
earth pressure E for 5 = 90 ® is given by the formula 

E = to tan* (+ 5 °- S/2 )_- 43 -A 

where H is the height of wall and w is the weight per unit 
volume of the back fill. 

The Pa ssive Earth Resi sta nce is the maximum force 
which the earlTTprism ABL, b, can just resist without 

being displaced by sliding upward on a certain plane of 
rupture BL, thus lifting the load P and bringing into action 
the resisting force F on the plane of rupture. 

The three forces Ep, P and F must be in equilibrium in 
the instant that any motion occurs. 

It should be noted that the minimum value of Ep governs, 
since the earth fill in resisting this force passively must 
do 80 without moving. The passive resistance Ep attains its 
minimum value when the angle 

According to Coulomb the horizontal component Ep , of 
the passive resistance, when Saeo**, is given by the formula 

Ep = u) H* f on^ (45“-*-S/2)-43-B 

Art. 44 Graphica l Solution, Wall without surcharge . The 
actual force representing the active earth pressure on the 
back of a wall is not the horizontal force E previously found, 
but a force E" of which E is the horizontal component. The 
force E' makes a certain angle , with the normal to the back 
of the wall, and this angle is now the angle of friction be¬ 
tween the earth fill and the wall surface, instead of the 
angle of internal friction of the fill material. The angle <f> 
is always smaller than S »nd depends on the nature and rough¬ 
ness of the wall as well.as on the fill material. 

Pig. 44-A will serve as a general picture of the several 
dimensional relations and forces involved in this and sucoeed- 
ing problems. The question of stability of the structural 
element or wall, not previously considered, will be taken up 
later. 

The method here given was developed by Poncelet and is 
applicable to practically all cases of walls inclined at any 
angle S less or greeter than 90°, and all cases with surcharge. 

The inclination of the back of the wall has a marked 
effect on the active earth ptessure E as given by Eq. 43-A 
which is true only for S =9o* or nearly so. When S is less 
than 90^ the earth pressure is less than for vertical walls, 
and when 8 is more than 90® the earth pressure is greater 
than that for vertical walls. 


91 










Gra phica I Construction according to Poncelet foR E*and Ijf, 


Fi^.44A. 

w =1 weight‘potuad8 per cu, ft. of back fill material. 

^ angle of repose or angle of Internal friction. 

<p = angle of friction between fill material and the back of 
the wall. 

S =. angle between the back of the wall and the horizontal. 

“VIT = the Tinknown angle which the surface of rupture makes with 
the horizontal. 

■i|/ = 4 - 5 "- 4 / 2 .=angle of surface of rupture for pasBire resist- 
ence. 

H = height of wall above grade in feet. 

P=» weight of earth prism ABL. 

&= weight of the wall section down to its foundation. 

The Poncelet construction proceeds as follows: At A lay 
off the known angle <§+ 4> and draw AC to the intersection C 
with the natural slope BS. Then describe a semi-circle with 
BS as diameter, and draw the line CC' perpendicular to BC. 

Now with B as a center and a radius BC' describe an arc to 
locate the point D. Finally draw DL//AC which locates the 
point L on the surface of rupture BL,thu8 determining the 
angle Y* • length DL is transferred on'to BS^ giving the 

triangle L D J, the area of which multiplied by w is the 
active earth pressure E' . The direction of E' makes the angle 
with the normal to the back of the wall and its point of 
application is at H/a above B. 


92 



















The same solution may he conducted hy descrihing a 
semi-circle on the diameter AB and proceeding as indicated 
by the dotted construction lines* 

According to the Poncelet solution, the surface of 
rupture divides the area ABDL into two equal areas making 
area ABL equal to the area LBD. 

The stability of the wall is found by combining E' 
with the weight G of the wall, into the resultant Reacting 
on the base of the wall. The vertical component N of R 
produces the soil pressure on the base. 

The horizontal component E of E' must be resisted by the 
passive resistance Ep in front of the toe, plus the resist¬ 
ance to sliding on the base which is Ntan4> . Allowing 
a factor of safety of two,the condition equation for safety 
against sliding becomes 

2E = EIp 4- N ton 4^ , where iAj-tNan^(4S®-*^ 2 )_44-A. 

Art. 46 Graphical Solution, walls with suraharge . 



When the surface of the ^ill is inclined at an angle ^ 
with tlie horizontal as in i’ig. iS-A,. the solution by Poncelet*s 
method is precisely as given in Art. 44. By noting the 
lettered points, which are identical in both problems, there 
will be no difficulty in following the solution as given 
first for the circle on BS and then for the circle on AB. 

The active earth pressure is found as E'=ujj«.arca LDJ. 

The same construction is applicable when the angle ^ is 
negative and the line AS falls below the horizontal. 


93 









Fig 0 shoire the case with sigoharge inolined at the 

angle ox repose mahlng where the point S is at in¬ 

finity# The method now takes on a special form# 

The natural surface of repose B D now becomes the sur¬ 
face of rupture and the solution consists simply in drawing 
the line AC with the known angle at A and then oon- 

struoting a triangle LDJ by drawing LDjfAC ^ and making 
nj - I n • The aotiTe earth pressure is E'=u>xarea LDJ. 



Fig# 45-C represents a broken surface lino which requires 
some adjustment before applying the Poncelet Solution. To 
accomplish this, draw the line BK and then draw AA'//BK making 
the triangles ABK and A' BK of equal areas^with A' on the 
prolongation of KS. The construction is now completed as 
before except that point A' now gorerns instead of A, although 
A'C'still r«m*ins parallel to AC. 

Art# 46 ^aotioal Wall Design Puts The vsurious kinds of 
walls treated herein may he divided into graTity walls (Figs, 
a to d) and reinforced concrete walls (Figs, e to h) and the 
latter may be of the cantilever, counterfort or buttressed 
type. See Figs. 46-A. 


94 










In determining the earth pressure E on the back of any 
wall, the extent of the backfill material and its physical 
properties must be known, also whether or not the backfill 
can assist by its own weight in resisting a part of the 
overturning effect due to the earth pressure. A surcharge 
of any kind, superimposed on the back-fill, will increase 
the earth pressure and must be considered in the formula for 
E. The enroharge is represented by a height h of fill 
material. 

The stability of a wall depends on two conditions; first, 
to resist the overturning moment of the active earth pressure, 
and also to resist the sliding on the base plana of the wall 
due to the horizontal component E of the active earth pressure. 

It is, therefore, necessary to determine this horizontal 
component E and also the vertical component N of all forces 
on the wall above its base plane, in testing the stability 
against overturning and sliding of any wall of given type or 
design. 

The earth press ure acting on the back of a wall, and 
tending to move the"waTl forward, is called the active eart h 
pressure E" and this has a horizontal component ^ to be 
evaluated presently. The passive resistance Ep is the 
force necessary to move a certain bank of earth against its 
natural sliding tendency and may be developed in front of a 
wall footing or in front of anchorages embedded in the earth. 
Passive resistance becomes important in dealing with a variety 
of problems to be considered later. 

The horizontal component E of the active earth pressure 
on vertical walls,according to^Rankine - Coulomb,la given 
by the formulas for 

walls with level fill E (45 - S/z) = ^ = K _ ^ 

walls wUh surcharge E = ^ H (H-r2h)tnn^(45tS/2^-^0^H-^2hjH=K{H+2h)H 3 

where w is the weight per ou. ft. of backfill and surcharge^ 
and § is the angle of repose of the material or angle of 

internal friction. 

See Pigs. 46-A for the several varieties of cases that 
may be treated with these formulas and consult Table 46-A 
for values of <5 , , Bnc\ 9^^ Yan'^(4s%S/2.), 

for various materials. 


95 






The r e su ltant diagonal earth pressure E_is a force 

mailing a certain angle of frictlbn"^*^with a normal to the back 
of the wall and having for its horizontal component the value 
of E from Formula 46-A* Values of ^ are given in Table 46-A. 

The point of appli cat i on of E is at r=H /3 above the base 
for liquias, and for materials listed in Table 46-A it is taken 
at r=V8 H when the backfill is level with the top of the wall. 
When there is surcharge, the point moves up to a position shown 
on the diagrams,with 

- Hh -h 3/8 __46-B 

■“ RTTFi 

The Resultant R acting on the b ase . The vertical loads 
consist of the weight of wall u Including the footing, and the 
weight P of the backfill when there is a heel. These are com¬ 
bined into the resultant Q representing the entire load on the 
base. The resultant R, found by combining £ and E graphically, 
represents the total effect of all forces acting on the support¬ 
ing base. The horizontal component of R is E, and the vertical 
component is N, acting on the base with an eccentricity e 
measured from the center of the base. 

The distribution of the vertical l oa d N on the base is 
found from ilavier’s law giving tKe following toe pressures, 
for dimensions in feet: 

lbs.per wh<fn £D/e , see 

■f « lbs.per sq.-fi., when t > D/g or when C < 

The unit toe pressure f should never exceed the allowable 
soil pressure as given below, and with this proviso, the 
factor of safety against overturning about the toe may be 
expressed as 

T = -3^---46-D 

where X is the norizontal distance from the toe to the resultant 
load Q, fig. (e). 

Safety against sliding on the base . The horizontal com- 
ponent E, of the earth pressure E' , may be resisted on the 
base of the wall by friction on the foundation between the 
concrete and the earth; by passive earth pressure developed 
in front of the toe; and by a key under the footing as in 
Fig. (f). Any, or all, of these resisting forces may be called 
into action in certain cases, and the wall design should pro¬ 
vide ample safety to resist a horizontal force of about £ E. 



96 











The frictional resistance on the base plane may be taken 
as Ntan<(>,and the passive earth resistance at the toe as 

Epi f t%Qn^(^5”+4>/2)-- 46.V. 

where t is the height of fill in front of the toe and ^ is 
the angle of friction between earth and concrete, Table 4C>-A. 
The resistance in front of the key may be taken at C.7 of the 
allowable soil pressure under the base. See also Problem 
47-C. 


97 





For values of «f>, e and k seetable 46/\ . wall sections analysed for i-ft. of length. 
ALLOWAB LE SO IL PRESSURE AT TOE OF WALL HYDROSTATIC UPLl FT ON WALLS 


foundation materials 

Ibsisq-tt 

foundation materials 

P 

lbs.s<^.ft. 


Mud,quicksand,silF, soft clay. 

piles 

Concrete on clay 

12.5 H 


Cloy .consis+ancy oi puffy, no visible wafer. 

1 o o o 

Masonry on cloy 

25 H 


Cloy,sfiff.moist,yielding sliqhfly undertWumb pressure. 

zooo 

Concrete on shale rock 

40 H 


Cloy, moist, dry enough to crumble. 

500 o 

Concrete on tine sand 

SO H 


Sond in natural situ and compact. 

4 o o o 

Concrete on Coarse sandorqravel 

62 H 


Sand and qravel .Compact as'm gravel pits. 

&000 

c-water level 



Sand clay and qrave 1 .compact, no visi ble wafer. 

GOOO 

Coarse qrovet, cemenfed with moist clay. 

G OO O 

a: ^ 

a i 
< 

u, ^ 


a 

H u 

Q 

f -iiniiiiiiii 

f .. 

1 

. ] 

Hard pan protected from water. 

7 0 0 0 

Stratified Clay protected from water. 

8 0 0 0 

Slate ond shole rock. 

1200 0 


;]Tj'||[TTrm|i|- 

Sandstone bed rock or better. 

30 ooo 

1 


D-Molitor, Jan.1930. 


98 

















































































































































































TABLE46A-HORtZONTAL COMPONENT OF LATERAL PRES5UREsC«^H^ tanVs-g^zVKti^. 


FILL MATERIAL 

Wfl&MT 

per 

cu.ft. 

LBS. 

Anqie 

a? 

repose 

Angle of friction 

4> 

1 Active. 

tan (45 

K=^0 

LBS 

tan 4* 
for 

Concret 

Passive 

tQn^(45®+f'2)= 

L 0p 

Concrete 

Surface 

Steel 

Plate 

Smooth 

Wood 

0 

Cam mon Earthed ry 

90 

01 

o 

0 

24® 

— 

— 

O. 3 33 

IS.O 

0.445 

! 3.00 

" •• mo»t 

lOS 

1 40 

30 



0.2 1 7 

1 1.4 

0.577 

459 

" •' v^et 

1 » 5 

20 

» 5 



0.49 O 

28.2 

0.268 

2.00 

Clay, dry 

J oo 

30 

24 



O. 3 33 

16.7 

0.445 

j 3.0 0 

•• moist 

1 1 o 

45 

30 


— 

O. 1 7 2 

9.4 

0.577 

5.82 

•t wet 

1 1 s 

20 

1 8 

_ 

- • 

1 0.49 0 

28.2 

0.325 

Z.OC 

Gravfil,cleon 1* 

1 04- 

48 

30 



O. 149 

7.7 

O.S7 7 

6.80 

Sand.qrovel.cloy 

1 30 

26 

20 



0.39 O 

25.3 

0.364 

2.56 

Sand .fine dry 

1 00 

3 I 

22 

15 


O. 3 20 

1 6.0 

0.404 

310 

" M moist 

1 1 0 

40 

30 

18 


0.2 1 7 

1 1.9 

O.S77 

4.59 

•• •• Wet 

130 

29 

1 O 


_ 

0.347 

22.6 

0.176 

2.88 

Cr. 1 1 me stone,fin e 

1 00 

35 

30 

20 

— 

0.27 1 

13.6 

0.S77 

3.70 

" “ •• Coant 

95 

45 

30 

20 

— 

O. J 7 1 

8.1 

0.S77 

5.82 

Broken stone 

) lo 

60 

30 



0.072 

4.0 

0.577 

13.95 

Coal, Anth. 

55 

27 

24 

1 6 


0.376 

1 0.3 

0.445 


« Bltum. 

50 

35 

30 

1 8 


0.27 1 

6.8 

0.57 7 


Coke. 

28 

45 

40 

25 


O. 1 7 1 

2.4 

0.6 39 


Cinders 

50 

38 

30 

30 


O.240 

6.0 

O.S7 7 


Ashcs(softcoalj 

42 

40 

30 

— 

_ 

0.2 1 7 

4.6 

O.S7 7 


Charcoal 

)2 

35 

24 



0.271 

1.6 

0.445 


Sowdu st 

1 O 

30 

20 


20 

0.3 33 

1.7 

0.364 


Hem. Iron ore 

I6S 

35 

30 


_ 

0.27 1 

22.4 

0.577 


Piq 1 ron.wifldijaist 

250 

SO 

— 

— 


0.1 33 

16.6 

— 


Cement 

88 

40 

30 



0.2 1 7 

96 

O.S7 7 


Salt CO ke 

65 

45 

30 

— 

— 

0.1 7 1 

5.6 

O.S7 7 


5odo ash 

75 

35 

24 


_ 

O.Z7I 

10.2 

0.445 


Cullet.peo size 

90 

-40 

30 


— 

0.2 1 7 

9.8 

0.577 


Malt 

33 

22 

24 



0.4 55 

7.5 

0,445 


Wheat 

49 

25 

24 

22 

20 

0.406 

99 

0.44 5 


Barley 

39 

27 

24 

20 

)8 

0.376 

7.4 

0.44S 


Oats 

28 

28 

25 

22 

20 

0.36 0 1 

S.O 

0.46 6 


Corn 

44 

28 

23 

20 

1 7 

0.3 66 

8.1 

0.424 


peas 

50 

25 

16 

14 

1 S 

0.406 

10.2 

0.287 


Beans 

46 

32 

24 

20 

1 8 

0.31 7 

73 

0.445 


Flaxseed 

41 

24 

22 

IS 

1 7 

0.4 1 5 

8.5 

0.4o4 


Water 

62.4 

O 

O 

— 

— 

1.00 0 

31.2 

0.0 























99 
































































































Art. 47 Problem 47-A . Design a reinforced concrete 
contilever wall with H = 17.5 ft., and surcharge h = 2.5 ft. 

Case (g) shown in Fig. 46-A. The analysis follows the 
description in Art. 46, and illustrates all minor details. 

The approximate width D of the base slab is found from the 
formula 

D = 0 . 85 (H-hK) ife* = 0-495 (H-t- h) 10.0 H. -_ 

The weight Q is the resultant of the several partial 
weights P-Hg-H g-»- and its position is found by taking 

moments about the heel. 

The earth pressure E is now found from Eq. 46-A as 
6570# with a lever arm r, from Eq. 46-E, and the resultant 
earth pressure E' is found graphically, as Indicated in 
Pig. 47-A. The resultant R is also found graphically by 
combining E' with Q. The vertical component N of R then 
gives the soil pressures by formula,46-C, all as given on 
the drawing. 

The table of stem moments is figured for heights of 4 
to 26 ft. from the formula 

M = Er = KH _ 47 -B, 

wherein H is treated as a variable, while h = 2.5 ft. con¬ 
stant. 

The required steel areas for different heights of stem 
are thus found and may be used for other examples involving 
similar data. The unit stresses used are f =650 lbs. sq. in. 
and 16 000 Iba. sq. in. ^ 

Problem 47-B is given to illustrate a case of cantilever 
wall with high surcharge using the formulae of Fig. 46-A,,case 
(h)# For the data given, the width of base slab is found 
from Eq. 47-A as D= o.85(H-f- = e.96-Ft., used 9.5 ft. 

The angle "ijr" =:45V§/2=6i-5o', determines the surface of 

rupture which is required to find the height h of equivalent 
surcharge. 

The toe pressure is found as j- - - T.7eo lbs. per sq.ft^ 

because 

The wall would not be safe against sliding without the 
key wall and the earth in front of the toe. 

Other features of the design are given on the drawing and 
require no further description. 


100 







3SI 


PR06LEM47A.REINFORCED CONCReXE CANTILEVER WALL. 



DATA FOR DESIGN. 

£t 6SO lbs., f = I 6 ooo Ibs.^ A ^ 

uj= too Ibi.j 3o® , 4 3 24-“ Soil pre<» 3 Soo a . 

-tan^^s"-?/!) = 16.4 t 

E *^(H-i-zh)HtQn^(4S'’-?/2)= KH(H-^2h).0) 

r — H h -t- Vs H ^ /-»A 

- H -i-ah — ------ 

M=rE = KH(Hhi-V'sH^’J s 41.7 HL6.2SH*forli*rs--(3) 

See table belo^^r -for sfeel recfuired m S+em. 

E 5 l6.66(l7.S^-2<2.S)t7.S = GS7o Ibs. 

y _ I2S «2S iTS*' _ 7. ocft. 

• 7.S -f Z * 2.S 

D*0,«5 (H-*-h)>|bilT4S-?/2)=0.'<WS(H + h)=r lolo“ 
Psia.c «4-9^»-loo s9lSo lbs X 2lSs zx,eso ii-lht. 

16 «. I.s K ISC'* » 3 6oo •• r S.6 « 20,1 S O •• •• 

^I.SkZ+-8k «asjlS0=l 9SO •• K S.O a 9,7 STo •• •• 

« 3.67 « 3*3 « loo a I 2 I o - X 8-25= IQ, OOP .. •• 

Totol Vert.Load Qt » IS 9 i o |bs« 3 . 94 * 62,750 ff.lbs. 
The graphic Composition of forces ifiVeS R,**/ith 
the ifdrticol Component No 194-00 lbs. ctcfini^ 
onfhe base of a point £= |.35" from the canter, 
where R prolonc^ec^ cut j the base. Then ihc Sotf 
pressure f iSto*a - 

Reinforcing steel aiid;f*369*a' 

The stem M^= E fT-t's)* 6570xS.S6 = 36SOO ft.lbs. ^^ 31.49 
Hc<?l = 8100 X 2*16 = 17 500 ff.lbs. 4^3 1.00 s^'*@S” 
Toe a 10750 * z- I ZlOx|.83»l928ot4.lb.,A^a J.ll*" 
Safety against Slietmg on the base , Ntan<^ > E. 

N ton4> = 19400 X o- A4S = 8 6 5o Safety j.jx 

Use IZ''xl8''key on base -for addi+ionol Safety. 

The factor of safety agonist over t urninq s ygNx = 2.53 ^ 
at the same time thaf+he toe pressure-f is not excessive. 
Total Concrete area* 37.0 S 4 .ft., 150 lbs. of bars per lin.ft. 


STEM moments for VARIOUS HEIGHTS H ANO h= 2.5Ft. AS PER EQ.3 


H 

M 

d 

A» 

H 

M 

d 

A$ 

H 

M 

d 

A, 




Steel area 




steel area 




Steel area 

ft. 

tt.lbs 

in 

sq.in. 

ft. 

ft. lbs 

in 

S4.in 

ft. 

ft.lbs 

in. 

34.in. 

■ 4- 

Joss 

15 

0.076 

12 

16 700 

11 

0.798 

20 

66,6 6 0 

Z7 

2.38 

6 

Z8S0 

I6.S 

0.1 8 1 

14- 

2S 290 

22.S 

l.l 1 

ZZ 

86 680 

28.5 

Z.9I 

8 

5866 

18 

0. 336 

16 

36 250 

Z4 

1.48 

Z4 

110, 50 0 

30 

3.5 1 

10 

10 420 

19.5 

0.54-4- 

1 8 

49.900 

2S.S 

1.90 

26 

1 38,000 

31.5 

4.16 


All values for I ft. length of wall. 


B.Molitor 
Jan ,1930. 


101 









































































Problem 47B. reinforced concrete: retaining wall. 



S.Mptifor 















































Pro blem 47-C Safety of wa lls aga inst S liding* The 
question of safety against sliding on the base, previously 
referred to in Art. 46, may not be the only manner of failure 
by sliding, as there may be planes of weakness below the wall 
base for which the stability condition is one of balance 
between the active earth pressure E, below the footing, 
against an equal or greater passive earth resistance Ep in 
front of the wall ss shown on Pig. 47-C. ^ 



wedge BPC from which the active earth pressure E, for the 
height t, is evaluated* The passive resistance Ep to the 
left of the wall and acting on the plane BC, is found for 
the height t, and must exceed E, , with a certain factor of 
safety . 

Eq. 46-A furnishes a value for E,^and Ep may be evaluated 
from Eq* 43-B, resulting in the following formulae: 


E, = ^t, (t,4.2:p)tnn^(45*-S/^) and 
Ep=: (45®-!-5/z^ acting at r = 3/8t above C 


47-C. 

47-1). 


103 










































Art, 48 Arches under faa Embanianents. 


The late ral eart h pressur es acting on a culvert or under¬ 
pass roadway may Ibe estimated according to the kankine-Coulomb 
theory although these pressures may be vastly altered by the 
method employed in placing the back-fill against and over the 
finished structure. 

Thus it may happen that the lateral pressure attains even 
greater intensity than that due to natural earth pressure, if 
the backfill is excessively tamped. Or, when little or no 
tamping is done, there may be a deficiency of lateral pressure 
and the structiire may be called upon to carry the vertical top ^ 
load without the assistance of the side pressure to develop 
full arch action. 

The total horizontal force d.ue to lateral earth pressure 
on one side may be greeter then the total force on the opposite 
side, a condition encountered when the surface of the fill is 
not horizontal. However, such inequality would adjust itself 
by a certain lateral displacement which would result in develop¬ 
ing a passive resistance at least equal to the active lateral 
pressure, less the frictional resistance of the structure on 
its basal support. 



104 




































AsBuming noraal earth pressures on both sides,and the 
weight of the earth over the top of the structure. Fig. 48-A 
illustrates the method of evaluating the earth pressures and 
loads P acting on the various voussoir sections. The condi¬ 
tion equation for the horizontal forces must be satisfied, 
observing that the value Ntan4> should^the absolute minimum. 
If necessary the values E® , * E^ , Eg may be increased 

to produce the balance between the horizontal forces. 

With the forces P established, the line of thrust may be 
constructed and the arch dimensions arrived at in the usual 
manner• 

The longitudinal force on a Culvert is difficult to 
evaluate witn any degree of accuracy. The following is to be 
regarded as a mere approximation. 

Let Pig. 48-B represent a longitudinal section of the 
structure taken through the fill ACDB. 




V = AT u; h cos. ^ . 
p saH uoh cos.^ 

R^rcsultant of V and p. 

Wssihc shaded area with ordinates t, and is 
maximum at F where the whole area from 
B to F represents the horizontal force H. 


Fi§.48B. 


105 



































The analysis is based on finding the resultant R for a 
length 4 I near the toe of the fill and this resultant will 
Intersect the axis of the dam in 0, which becomes the common 
pole for all resultants R. The horizontal components of the 
several forces R are then plotted as ordinantes to the base 
AB, and the area of this curve, on one side of the center F, 
represents the total horizontal force H. 

This method is given by Dr. H. Krey (Erddruck, Erdwider- 
atand) p. 170, with the reminder that it cannot be theoretically 
substantiated though it may serve a useful purpose in the 
absence of a better method. 

To analyse a given arch ring by line of thrust method , when 
the externally applied forces have been determined as previously 
described. Referring to Fig. 48C, proceed as follows: 

I. Divide the arch ring into a suitable number of 

voussoirs,or hypothetical arch stones, so as to insure a Joint 
at the crown and one at each support. The location of the 
voussoirs is a matter of convenience. Then find the external 
forces acting on the several voussoirs and combine these with 
the vertical weights g to obtain the total forces p^ to p to 
be carried by the arcH ring. ' 

II. Construct a force polygon by adding graphically the 

several forces p^ to p^ in their proper order, true directions 

and magnitudes,"using”any convenient scale of forces. The 
resultants P, and of. the two groups of applied forces to 
the left and right of the crown Joint c, are thus found in 
direction and magnitude. 

A point on the line of action of each of the two forces, 

as d, for P, and d^ for P^, is found by drawing any trial 

equilibrium polygon through the several forces p^ to p for 
any convenient pole O', with pole distance t. The points d, 
and dj are the intersections of the closing rays r, and r^, 
with t, and fix the positions of the resultants P, and P^. 

This much of the drawing may be made in ink so as to pemlt 
of erasing subsequent construction lines drawn in pencil for 
the purpose of locating the most probable line of thrust . 

III. An equilibrium polygon can now be drawn through 
any three assumed points c, a and b of the crown and springing 
Joints, and as a first trial it is well to choose the axial 
points of these three Joints as giving the best Indication 

of such shifts as may become desirable for subsequent trials. 

The line of thrust corresponding to any equilibrium polygon 
is merely another broken line Joining the points of inter¬ 
section of the resultant rays with the voussolr Joints. 

To draw an equilibrium polygon through the three axial 
points a-c-b, it is necessary to first find the crown thrust T 
acting iit”c. This thrust, is horizontal for symmetrical load- 


106 






107 


























ing, while for unsynunetrlcal loadlng^as in the present case, 
its direction is found in the force polygon as follows: Since 
the points a-p-b, if they are to be on the line of thrust, 
must necessarily be points of zero moments,_therefore, the 
reactions of the force P, on the half arch ap must be Kj, 
(through b and £) and K, (through a and n, )_, Llkewlae the 
reactions of the force on the half arch must be K 3 
and But the crown thrust T is the resultant of and 
and may be found in the force polygon, by resolving P, into 
K, and Ki,also P^ into K, and K^jand then finding the re¬ 
sultant of and Ks equal tq_T=OH. By drawing T through 
the point c and parallel to OH in the force polygon, the 
points m, and m^ are located, and the reactions R, and 
are fully determined. Finally with Q as the pole, draw the 
equillbri-um polygon through the axial points a-c-b, (not shown) 
and then connect the points of intersection of this equilibrium 
polygon with the several voussolr joints, to obtain the line of 
thrust as shown by a light dotted line. In like manner other 
lines of thrust may be constructed through any three points 
of the crown and springing joints. 

By inspection it is seen that the line of thrust just 
found, passes close to the intrades at joint 6 , which appears 
to be a critical section, while in the left half of the arch 
it closely follows the arch axis. This indicates that the 
true line of thrust must come closer to the arch axis over 
the rlglit half of the arch, while on the left half the new 
line must depart more from the axis in order that the internal 
work of deformation may be reduced to a minimum, and thus more 
nearly equalize the stresses at all points of the arch ring. 

For these reasons we construct a new equilibrium polygon 
and resulting line of thrust , choosing the three points as 
follows: Move the points a and b to the right,or middle third 
points of these sections, and move the point £ up 0.17 ft. to 
the outer third point. By repeating the above method we locate 
the new pole 0 ,, and the heavy dotted line as the new line of 
thrust. The stresses on any arch section as joint 1 , are found 
graphically as shown on Fig. 48C,or they may be computed by 
Eqs. 46-C. 

Now that a new line of thrust has boon found for the 
given case of loading,revealing critical sections at a, 1 , c, 

§ and b, how is one to know that the critical stresses found . 
for those sections are actually the minimum obtainable values? 
The criteria will now be given for locating the most probable 
line of thrust , such that the critical stresses simultaneously 
attain min iTmim values. This requires finding the most probable 
line of thrust from among all the possible ones which might be 
drawn. 


108 








Criteria for locating the moat probable line of thrust 
have been proposed by Gerstner 1831, Moseley 1837, Hagen 1844, 
Bauernfelnd 1846, Schwedler 1859, Culmann 1866, Durand-Clayo 
1867, Winkler 1867, Helnzerllng 1869, Von Ott 1870, Belpalre 
1877, Muller-Breslau 1886, J. Weyrauch 1897, Landsberg 1901. 

According to Menabrea’s Law of least work, the statement 
regarding the location of the most probable line of thrust In 
an arch ring, would be that It must correspond to a minimum 
work of deformation. The actual Internal work due to bending 
moments M and direct stresses N, by Ea. 15K, p. 42, Kinetic 
Theory of Eng. Structures, Is 



48A 


Since M = Ne> where e Is the eccentricity of N, measured 
from the arch axis, thereforeEc^, 48A may be written 



4-8 B 


Referring to the force polygon Fig. 48C, It will be seen 
that the resultants R,and hence their normal components N, 
acting on the severaT voussolr sections, are only slightly 
affected by small shifts In the pole 0, hence the second term 
In Ecjf. 48B may be regarded as practically constant so long as 
the arch ring and Its loading remain unchanged. Also, the 
products e* In the moment function will depend essentially 
on the variable e"*. Therefore, the Internal work due to moments 
will become minimum when _/^e*dx Is minimum,and since we are not 
Interested In the numerical value of W but only In the conditions 
which reduce Its value to a minimum, Therefore, the second term 
disappears as a governing condition. 

Hence, the condition which reduces the actual work of 
deformation to a minimum Is that ^e^dx - min., and If the 
voussolrs are spaced at equal Intervals making dx constant, 
then the condition Is simply thatSe*= min. Stated In words, 
the most probable line of thrust Is the one which reduces to 

a minimum, the residual departure from the arch axis ( center 

line) according to the method of least squaresT 

For example. If a line of thrust can be made to coincide 
with the arch center line, then this thrust line will be the 
most probable one of all the possible lines for the given case 
of loads, as the valueSe* will then be zero. It was for this 
reason that the first equilibrium polygon was passed through 
the three axial points a-c-b in Fig. 48C, so as to reveal the 
character of our problem and then decide what shifts to make 
in order to comply with the above criterion. Thus,for the 
line of thrust drawn through the three axial points, with 
values of e In inches, we obtain 85.67, while for the 

heavy line of thrust through the middle third points40.72. 

A line of thrust drawn through a and the outer third points at 
c and b (not shown) gives^6.4 which may be taken as 
Sufficiently close for a minimum value. 


109 










TTiere Is a further condition which must be satisfied 
in order to balance the positive and negative values of e. 

It is that the line of thrust (or equilibrium polygon) must 
intersect the arch axis in at least three points, and when 
the arch and its loading are symmetrical there must be at 
least four such points of intersection. Our final line of 
thrust has three axial Intersections and complies with this 
requirement. 

The method here outlined for the analysis of arches 
should always be employed when dealing with short span, 
semi-circular or high arches, having rather thick rings and 
subjected to non-parallel loads, as for arches under high 
earth fills. 

All methods for the evaluation of the three redundant 
conditions in arches of the kind here treated , are more or 
less speculative, but yield results which are quite within 
knowable limits, especially when the heterogeneous character 
of masonry is considered. Whether we locate the line of 
thrust by the method of least work, by the elastic theory, 
or by a few trials, observing the criteria above stated, 
makes very little difference except in labor involved, and 
certainly the least laborious solution is the most acceptable. 

The above method was devised by the writer for the 
analysis of the Gatun Lock walls of the Panama Canal in 1907, 
as illustrated in Fig. 48D. The.lettering corresponds to 
that employed in Pig. 48C, so that the description need not 
be repeated. 

When dealing with large flat arch ribs, or barrel arches 
with thin concentric rings, carrying vertical loads, then the 
theory of elasticity affords the best method of stress analysis. 
See the writers Chap. 15, Kinetic Theory of Eng. Structures, 
for an exhaustive treatise on this subject. 


110 





David Molitof. 
1907. 


Ill 
























































CHAP. 10 - 3HEET PILING PP.OLLIXS. 


Art. 49. Sheet pil es in level e ar th w ith horizo ntal l’orce_ 
acting at the top as shown Fig. 49-A.” All loads per 1 ft. oT 
piling. 



Let w s weight of earthy pounds 
per cu. ft. 

f z passive resistance per 
sq. ft. at ground level, 
which should not exceed 

u;t 0p tot tan*^ S/z j 

for a factor of safety of two 
e = pounds per sq. ft. 
pressure at the bottom of 
sheet pile, which likewise 
should not exceed the allow¬ 
able valueu;t 0p. 

Two condition equations may 
be written as follows : ^ P =:0 

and S = o. 

Hence from Fig. 49-A, 

s: P= P-P, 4- ==o 

SM.= P(a+t)_^P,.tP^ = 0. 


aubstitutlng values for P, and from the figure, then 


P-S+^=o and P(a*t)_^+^io . 49-., 

The solution of these equations gives 
f = ~ 2t) and ...49-B. 


which values should not exceed the passive resistance wt© 
or -^wtep when allowing a factor of safety two. 

The maximum bending in the sheet pile occurs at about o.zir 
below B. 


EXAMPLE 49-A. For wet sand with w«130 lbs. per cu. ft., and 
§ = 29* 0p= tan* (45V 2.8ft, then With sheet piling 26 ft. long 

driven 10 ft. into the ground, Ei^s.49-B give 

f ^ ^ ^ ^ I6 4'2aIo^=i. 36P, ande ^ixi6 + io^r= j.ie>P. 

These values should not exceed jwt0p= LLQ x iox 2 . 8 &= ts&z 
lbs. per sq. ft., hence a safe value for P should not exceed 


112 



















f^.36 = • = 1^7 0 pounds. The maximum bending moment on the 

pile, per it. of wall,will be about P(a+ o.zt]= 1370x 18=24,600 ft, lbs. 
to develop the available resistance lor the 10 ft. penetration 
with 8 safety of two. 


Art. 50 . Sheet piling b ack filled wit h ea r th on one s ide 
and a uniform live loa d c y p er sq. ft. on t he fill . 



Let w = wgt. of earth, lbs. 
per cu. ft. 

0 = +an^(45 - S/ 2 ). 

P, , Pj , Pj end P^ are 
earth pressures as indicated 
in Pig. 50-A. 

f = passive resistance, 
lbs. per sq. ft. at ground 
level which should not exceed 

^w+0p for 0j,= tan^(45‘’+S*/2 ) . 

e = passive resistance 
lbs. per sq. ft. at bottom 
of sheet pile. 

The condition equations for 
sP = o, and SMg=o may then 
be written as follows: 


and 


P 

I 




-60-; 


Fi^.SOA. 

P-+-P. — R = 0 ,or VO0 cwH0-*-i+_ a = 0 
i * ^ z. 22 

> a? 0H * tj -h cwH0 (t- — ^ =.0 . 

where c=ty^=-o.6ot approximately. 

Equations 50-A, with numerical values for any given problem, 
will serve to find the unit soil pressures f and e for an assumed 
value of t. These soil pressures should not exceed the passive 
resist€mce 4vot0p for a factor of safety of two. 

The point of max. moment in the sheet piling is at about 
i/3 below B. 

Substituting the approx, value of c=:o.&t into 50-A, 
and solving for f and e, gives 

j- — -f I.32IA)H0 

e = fH+t) -h o. izvoH© 


} 


.50-B. 


113 






























Example 50~A. H- 16 ft., t « 10 ft., w = 90#, 30**, 0 = 0.333, 

0p=3.oo, then Eqe. 50-B give f r 3398 lbs., sq. ft. and e r 2064 lbs. 
sq. ft. 'Fhe passive resistance is only u)t 0p= 2700 Ib^.^ without 
any allowance for safety. This means that t must be made greater. 
Max. moment = 8.67 x 3840 = 33^290 ft. lbs. 


Art. 51 . The effect of Relieving Platforms in reducing the 
active pressure on the face wall is important and may be briefly 
illustrated without regard to the manner in which the wall is 
supported. The questions relating to wall support are taken 

up separately in dealing 
with back anchors and 
sheet pile penetration. 

The platform may be 
supported on piles and 
takes weight off the fill 
beneath the platform, 
thus relieving the hori¬ 
zontal pressure against 
the front wall in a manner 
indicated by Pig. 51-A. 

The pressure area for 
the height h above the 
platform has a base vA)0h 
lbs. per sq. ft. and the 
same line is extended 
down to B where the value 
becomes toOH lbs. per 
sq. ft. Drawing the 
natural slope line through 
D to P, and the plane of 
ruptxire through D to E 
determines the points 
where the pressure changes, 
giving the shaded area as 



Fi|,.5l A. 


the total pressure area active against the face wall 


114 



























Art, 5£ , Sheet piling back filled and with baot anchor. 

Condition equations♦ 



u} = weighi oi I cu.-ft. Earth fill. 
^ = angle of repose of earth- 

Fi^.52A. 


P-^ = o 


to find 
assumed 


f and 

t. 


P for an 


A certain factor of safety 
should be provided against 
failure at the bottom so 
that the passive resistance 

b=nf, may have a factor of 
safety n s i*>f6 p , which 

should be about two* 

- (t+ H) * A 

3 ]"■ 


giving the value 


5£-A. 




- 5E-B. 


Example 52»A . 16 ft., t= 8 ft. u>* iis lbs , e=o.io 8 , g= 2 oft., 

which values give mo --l^s* eq* ft 

The total passive resistance is b= vA>t 0p* ns «. 3 . 24 -s = 29ao 

lbs. per sq. ft. which is not quite 2 f. 

The pull on the anchor P= (t + H)^_ tl = 3izo lbs. per 
ft. of wall. ^ 

To find the point of max. moment, write the moment equation 
for any point x below A and make 3M- o to find x. thus = 

0 K 

M •=* p (x- a) — , and - P — «i5L@ x.’’ = O 

making x li.zv ft., and M^=P(j3.a7-4-)-35.4z = »s 200 ft.lbs. 


115 























Art, 53 The Combined Action of Earth and Water on Sheet 
Piling. 

This problem is usually encountered in designing Quay walls 
or marginal wharves and involves many uncertainties, the nature 
and magnitude of which cannot be appraised with accuracy. 


Case I. Coarse Material for Back»fill« The physical aspects 
of this problem will Tirst be discussed by assuming a line of 
sheet piling with water on one side and rip rap or crushed stone 
on the other side, both at the same surface level as shown in 
Pig. 53-A. 

Let w = weight of one cu. ft. water 



= 62.4 lbs. 

w, I weight in air of 1 cu. ft. 
dry fill material. 

^ “ volume of solids in 1 cu. ft. 
of fill material. 

I volume of voids in 1 cu. 
ft. of fill material. 

0^ = tan ^ ( 4-5 - the Rankine 
co-efficient in the formula for earth 
pressure, where is the angle of 
repose of the wet materiel represent¬ 
ing the maximum effectiveness of the 
back-fill. 


The water pressure on a square foot, located h ft. below 
the water level, will be wh. This will be balanced’ by an equal 
and opposite hydrostatic pressure on the side of the fill, a 
condition which must prevail if the voids in the fill are 
sufficiently open to permit the water to circulate and wet the 
entire surface of sheet piling. In addition to the water 
pressure, there will be the effect of the back-fill, weighing 
(lo,_£,vo ) pounds per cu. ft. in water, and exerting a horizontal 
pressure -eiA> )e^H, in pounds per sq. ft., due to the 

submerged weight of the wet material. 

The total unit pressure on the filled side, at a depth h 
below water level, thus becomes ~ 


p = w H -t- e 



IV, 0^ H 4 - [ I — ^ ^ H 


53-A. 


116 















This is the usual method of combining the hydrostatic 
pressure with the earth pressure and represents the combined 
effect due to full hydrostatic head increased by the pressure 
of the wet material acting under its submerged weight. 

By assigning numerical values to the quantities in Eq. 53-A 
for the case of rip rap back fill of limestone,weighing 98 lbs, 
per cu, ft, and containing 3o;b voids, making £.= 0,65, then 
for an angle of repose 40°, the value 0^= tan*(4-5 SA) - 0 . 2.17 
giving ' “ ^ 

pzs 9g X 0.21 7 H + (^ 1 - 0.65x0.2 I 7j 62.-f H = 74. eH lb». per s^-rt. 

This is 20^ greater than the water pressure in front of the 
wall. The triangular areas in Fig, 53-A, thus represent the 
total pressures on the two sides of the^ sheet piling. On the 
water side the total pressure is 62,4_H' acting at H /3 above 

the bottom. On the side of the back fill the total pressure 
is P = also acting at H /3 above the bottom, 

2 * 2 -’ 

Case II Fine Material for Back Fill, The distinction 
which is here made between coarse and fine mat erial for back 
fill,requires some explanation without which^the purpose 
of our classification would not be clear. 

In speaking of coarse material the term was intended to 
apply only to rip-rap, crushed stone, gravel, or coarse sand, 
the voids of which are sufficiently open to allow water to 
circulate freely between the interstices and to run out or 
drain off whenever the material ceases to be submerged. 

Fine material on the other hand may be described as 
having voids of such small dimensions that water cannot cir¬ 
culate between the particles and would be retained more, or 
less, even when the material is removed for a short time 
from the submerged state. Fine silty sand, clay, marl and 
mixtures of these with each other, or with coarser material, 
would be included under fine material and may retain from ten 
to thirty-five percent of water filling the voids. 

The behavior of fine material, as here described, differs 
radically from that of the coarse variety,irrespective of the 
percentage of voids possessed by either class. 

In the case of fine material a new physical law mani¬ 
fests itself in the property known as capillary attraction, 
implying a specific attraction between the water and the 
solid particles which are in such close proximity as to hold 
the water sufficiently to overcome the force of gravity. Such 
a material, once saturated, will lose little weight simply 
by drainage, and most of the retained water can be removed 
only by evaporation or pressure. 


117 







Therefore, when the hack fill consists of fine material 
as here described, the above analysis of pressure given by 
Eq. 53-A, no longer applies and new assumptions become- 
'necessary. In no event can the full hydrostatic head now 
exert itself and we might deal with the problem as one of 
wet earth pressure which according to the Rankine formula 
might approximate hydrostatic pressure as a maximum. How¬ 
ever, this may not be sufficiently safe for a completely 
saturated material such as we are now considering, and it 
would appear wise to estimate the maximum possible pressure 
value which might be encountered. 

Such a maximum pressure value could be made up of the 
increment due to the dry weight acting with the highest 
Rankine coefficient for wet material, plus the hydrostatic 
pressure due only to the water filling the voids. For ex¬ 
ample take a fine silt having a dry weight = 100 lbs. per 

cu. ft., and (l-6 )= 0.40, with an angle of repose §= 26® 
for the same material when thoroughly wet. Then 6^=tan^(45- 
ss 0 .^ 06 , and the combined pressure would thus become: 

Pressure due to silt «.w,0^K=^loOxO.4O6H=:4o.aH. 

Pressure due to water in voids £)wH = o.4k62.4-H= ; 25 . 0 H. 

Total pressure of fill material = p =i6S.6H. 

which is slightly more than full hydrostatic pressure of 

62.4 U. 

The total pressure thus found may be expressed as a 
formula thus p=uj, (i-aJwH-63-B. 

The weight of this material when saturated, would be 
100 + 26 = 126 lbs. p. cu. ft. and would exert a pressure of 
126 X 0*406 M= 60.76 H pounds per eq. ft. according to 
Rankine*8 formula. This value is considerably under that due 
to water alone and is undoubtedly too low for saturated 
material. The pressure given by Eq. 63-B, would be 66.6 H, 
and this accords well with observations. 

In accordance with the reasoning employed in arriving 
at Eq. 63-B, the solution of a sheet piling problem may be 
outlined as shown in Pig. 63-B. Quantities are per foot 
length of wall. 


118 






6;'^.4 lbs. 

dry weight of fill materi¬ 
al 

wet weight of fill materi¬ 
al 

volume of solids con¬ 
tained in 1 cu. ft. fill 
material, making volume 
of void# s l-g.« 0.25 to 
0,40 cu. ft. 
tan^ for dry 

mater ial. 

tan'i ( 4 . 5 for wet 
material. 

tan^ S'/z) = coeffi¬ 

cient for passive re¬ 
sistance for wet materi¬ 
al. 

pull on-anchor rod. 

P,, , P, , P^ are pressures 
in pounds for the pressure 
areas as indicated, and 
acting at the centers of 
gravity of the respec¬ 
tive areas, with lever 

arms r above the base KZ. Finally R * lbs. per sq. ft. 

passive resistance at the bottom of the sheet piling, making 
the triangular area ^ = Py represent the unbalanced force 
required to hold back the toe. The maximum value of R = u;^ept 
should never be developed to more than one-half this value, so 
as to Insure a factor of safety of at least two against failure 
by kicking out at the bottom. 

In the above description of the problem everything may be 
regarded as known or assumed, except the anchor pull P and the 
bottom pressure R, and these may be regarded as the forces 
necessary to supply the conditions for equilibrium of the beam 
ABC. To find these two unknowns we may write two condition 
equations as follows: 

Sum of hor. force8=i0 , or p- p _ P, 4 - P, ^ P, + P_ =0 

' ® 1..53-C. 

S)un of Mom. about C=o, or ?6-?r-Pr,+Pri-Pr.^ftr,= 0 J 

Noting that f|= and Si* we obtain the follow¬ 
ing: ^ ® 


Flg.53B. 


119 

















53-D 



which may he solved for R and P to obtain: 

P- P+P-P_P _ Rt 

The method Just outlined will now be illustrated by solving 
a complete problem* 


} 


-53-E 


120 




Example 53A. Sheet P^unG: Revetment with back anchor . 

The back-fin is fine Sand, assumed as dry abov'e wa-f-er level and Saiurafed be low water. 
The loads are per linear foot of piling , Se.e F/g.53A for dimensions 

DATA. 



R*—^,j_r8,0l5-^(S40x3S-h 4l,630>i 11.8- IS.OOOx 17-16, J5 5x4.25j -697 
^ 3x31 L -I 

Ps: P 4 . R - - P._ Et = 8.015- 4.5x 8^7 = 3e80 Ibs. 

I P *T ^ 

This j^i^es for E = 4.5 R * 4 ,036 lbs as against an avoilable value of ***tPp^ = 

125 X 2.44 K 5.'’= 12,4 54 lbs., representing a factor of safety of ii±§±_ = 3 .1 - 

The bending moment on-fhe sHeetpilin^ will now be found by writing the rnoment 
•for any point x feet below Water level. Thus: 

l^(x4-2^— &5.C _ I SO x ^ j and substituting fbe obove values oi PandTf ^ 

get M^s 10.4 x*+ 39S0X- 7960 - 540 X — lo®o — 10 . 9 K*— 9o x* , which simplifies into 

M^ = -0.5X^- 90X^4- 3440X —9040 -------53A. 

To find the value of x for which is maximum , eguofe the first differential derivative 
of Eg.53A to zero ond Solve for x . This gives : 

^ hix — _ 1. s x^— I so X -h 3440 = 0 , from which x = 4 -16.7-ft., and wi-lh this value of x, 

d X 

Eg.S3A gives maximum 2 0,96o ft.lbs. The moments forx = o, x = 2 , x= Io and xs. 24 ff. 

were also found fro m Eg. 53 A, and are plotted in Fi^.S3C to show the enti rc moment curve. 


121 


































Art« 64 , Resistanoe developed by baok anchor Let the 
plate^^6 represent a continuous anchor, sahmerged below the 
ground as shown in Fig. 54*A» All foroes are per unit length 
of anchor plate* 



Fi§.54A. 

Let ? represent the safe resistance of the anchor per 
foot of length which may result from the passive resistance 
of the earth in front of the plate* less the active earth 
pressure E 2iv0(T-t*} on the back of the plate, neglecting the 
active pressures on each side of the plane AB » t* 

Introducing a factor of safety n into the passive re- 
sistanoe , the following equation may be written 

for a length of one foot of a continuous anchor« 

P= w(^_e)(T^-t^)---- 

where P is assumed to act at the middle of plate BC* 

When the force P approaches the condition for failure 
then the prism PPBC must be moved forward and upward on the 
plane of rupture DC,while the prism FSCB would slide on the 
planes of rupture CS and FB* 

For an ^ohor plate of length b , the passive resistance 
becomes relatively ^eater^owing to the lateral spread of 
the passive earth resistance at the two ends* 

The safe pull on an anchor plate of area b(T*t) is 

P=wkQpT^- 'ikSf''"^siJiLTHan.?__ _ 

<2nP ^n/jBn ^ o4-B 


When the clear distance between the edges of such anchor 
plates has the value d,or lessees given by 


^ _ ZuoOLT^tan^ 


64-C. 


then the effectiveness of the Isolated anchors becomes equal 
to that obtainable by a continuous plate as per Eq, 64-»A, 


122 















Example 54-»A . Supposing the anchor plate of height 6 ft. 
is submerged 5 ft. below the surface in ordinary earth with 
w s 100 lbs. per cu. ft. and s = 30°, what is the safe resist¬ 
ance per ft. length of anchor? This makes t - 5 ft., T = 11 ft., 
0 =taM^(45 -S/ 2 ,) o. 333 and 0p = tan^(4s+5/2)= 3.0 0, Assuming a 
factor of safety n r 2, then Eq. 54-A gives : 



How suppose we have isolated anchor plates of length b i 10 ft. 
find the safe pull for one such plate. With the same data as 
above and n s 2, then from Eq. 54-B for* L= T cot{4-5‘*-S/2) = l9.os ft., 

we have: ^ 



I y 


or &o,o60 1bs. whioh represents say 8006 lbs. per foot of length 
as against 5620 lbs. previously found for a continuous plate. 

The distance apart d,which such plates can be spaced with¬ 
out loss of effectiveness,is given by Eq. 64-C as 

d = >00x0.333 X 19 . 0 s X > 2.1 KO.S 77 _ f 3 ^ 

.= 3[3.0, 111 -0.533 «2S] 

While these values are necessarily quite approximate the 
above formulas undoubtedly lead to far more rational designs 
than are commonly made. 

Art. 55 Length of Anchor Rods and the location of Anchor 
plates will now be considered, in order that the anchorage 
may function properly in supporting a wall subjected to earth 
pressure. 

There may be three possible cases which would govern the 
effectiveness of anchor plates as follows: 

1. Anchors located between the wall and the plane of rupture. . 

2. Anchors located between the plane of rupture and the natural 
slope. 

3. Anchors located back of the natural slope. 

Anchors located in the pressure wedge between the wall and 
the plane of rupture cannot have any practical value and deserve 
no fxirther consideration.^ 

Anchors located at c' , beteeen the surface of rupture BD 
and the natural slope BF, cannot in themselves offer any support 
to a wall AB, since the back-fill may slide on a surface which 


123 











is steeper then the natural 
slope without developing any 
appreciable passive resist¬ 
ance. See Pig. 55-A. 

However, when the wall is 
otherwise supported as by 
driving the sheet piling 
into the ground at B, or in 
case of a masonry wall ade¬ 
quately supported on a good 
foundation, there may be a 
divided responsibility be¬ 
tween such supports and the 
back-anchor even when each 
alone would be inadcqioate to 
furnish the necessary sta¬ 
bility. It is therefore, not 
advisable to place much dependence on anchors of this class. 

The only safe anchor is one which is located behind the 
natural slope as shown at C Fig. 65-A, and the distance back 
from the wall should be such that the intersection of the two 
planes of rupture BBand DC as at D , falls practically on the 
surface AS,or only slightly below. 

The formulas previously given for safe pull on the anchor 
plate are then applicable. 

For level earth sxirface and vertical sheet piling the 
length of anchor should be not less than 

1 = Htan(4S-- S/jj + Ttan(«-+S/ 2 )-66-A. 



134 








CHAP> 11> WAVS PRESSURE, SEA WALLS, BREAKWATERS 


t 


Art. 56 On the Nature of the Problem. Sea walls and 
Breakwaters are exposed to the action of waves generated by 
high winds, sweeping orer a considerable expanse of open water. 

The size of a wave for any particular locality depends on the 
velocity of the wind, duration of the storm, depth of water, 
and the greatest distance over which the wind can act, pro¬ 
vided the water is of sufficient depth for wave formation. 

The distance from the windward shore is called the fetch. 

Assuming a constant direction and velocity of wind, the 
height of a wave will Increase from zero at the windward shore 
to some limiting height which the assumed wind velocity will 
just maintain. When the fetch is sufficiently great and the 
depth of water is ample, the wave will attain the limit of 
height before reaching the leeward shore, continuing with a 
constant height until some resistance is encountered in shallow¬ 
er water or against a structure of soms sort. 

AS a wave increases in size according to the distance out 
from the windward shore, its length also increases and so does 
its velocity of propagation. When the balance is finally 
attained between the wind velocity and the wave dimensions, 
then the velocity of the wave no longer increases, but may 
continue without reduction for long distances even after the 
wind has subsided. 

Assuming for the present that the height of a wave may 
be estimated for a given fetch and wind velocity, as established 
for a given location where it is proposed to construct a sea 
wall or breakwater, then the next step is to evaluate the 
probable mass and velocity of such a deep water wave and finally 
trace its retardation and diminished dimensions as it advances 
into shallower water previous to striking the wall where its moving 
mass is stopped. 

In case the wall is so situated that the maximum wave can¬ 
not strike normally, then there will be a further reduction 
in resolving the wave force normally to the face of the wall. 

Having finally decided on the probable size of wave which 
may be expected to arrive in front of a wall, we next estimate 
the height to which this wave will be piled up when completely 
obstructed, and then decide on the height to which the wall 
is to be built. If the wall is not carried up to the full 
height of the completely obstructed wave, then a portion of 
the wave will go over the top of the wall and the wave will be 
only partially obstructed. 

When a wave is partially or completely stopped by a wall, 
then a portion, or all,of its energy is converted into a 
dynamic force, which in turn represents the static equivalent 


t This Chapter was publisher:/ jn Transactions,/)m.5oc.C.E'.,l936. 


125 










of the expended energy. The wall structure must be capable of 
resisting this static equivalent without exceeding certain 
requirements of structural stability and safety. 

The various phases of the problem as above outlined will 
now be examined in detail, to show how each element may be 
quantitatively evaluated in a logical sequence. It should be 
remembered however, that all formulas employed in this con¬ 
nection are based on observations which are in themselves 
quite approximate, while a theoretical basis iff only remotely 
possible. 

Most of the data employed herein are taken from 
Professional Papers No. 31 of the Corps of Engineers, U.S. 

Army, on ’’Wave Action in relation to Engineering Structures", 
by D.D. Gaillard, Capt. Corps of Engrs., 1904. The applica¬ 
tion of Capt. Gaillard*8 observations, supplimented by others 
made in 1915 at Toronto, Ont., was developed by the author in 
connection with the Toronto Harbor work. 

Art. 57 Height of Waves in Terms of Wind V elocit y and 
Fetch. For a c-iven wind velocity in miles per hour, and a fetch 
D in statute miles, the height,h in feet,of,a wave, may be 
estimated from the formulas 


h = o,i 7 'jvD for values of D greater than 20 miles 

h=0.17^76for ^ less than 20 miles 



which are based on the formulas of Thomas Stevenson, after 
introducing wind velocity as a variable, and employing 
statute miles instead of nautical miles. 

These formulas apply only to inland lakes where the 
fetch is not likely to exceed the distance subjected to a 
violent wind in a single di ection. 

With few exceptions ocean waves rarely exceed 45 feet, 
corresponding to a wind velocity of 75 miles per hour and a 
fetch of about 935 miles,which represents an unusual combina¬ 
tion of circumstances. Ocean storms are generally more or 
less local, and do not cover more than 50 to 100 miles. Since 
the oceans are of much greater extent than any possible storm, 
the fetch is restricted to the storm area which may be any¬ 
thing from a few miles to say five hundred miles, and no 
formula can be made to apply to such indeterminate conditions. 
Hence, the only reliable data relative to the height of ocean 
waves must be collected by direct observations for any given 
locality. An extensive collection of ocean wave observations 
is given on p. 76 of Capt. Gaillard*s treatise. 


126 








Art. 58 Len^rth of W ave in terms of heig ht an d wind 
velocit y. The maximum wave has the"minimum length ratio and 
as the wind subsides the height diminishes and the length 
ratio increases. Fig. 58-A gives certain dimensions for waves 
which will be used throughout the present chapter. 



t« the period of the wave, or time in seconds for the 
crest to travel L ft. 

Vs L/t = the velocity of wave propagation in feet per sec. 

Fig. 58-A. 

a. The ratio L/h varies between rather wide limits 
depending -on the velocity of the wind V, the duration of a 
storm, and the depth of water in which‘the wave is formed. 
According to observation by Capt. Gaillard in fresh and 
relatively shallow water, this ratio varies between the limits 
9.1 and 15. A formula here proposed, expresses this ratio in 
terras of wind velocity V as follows: 

L ^ 840 or L=84oJl-58-A 

h V V 

This gives the following values for various wind velocities. 


Vs 75 70 es 

60 

55 

50 

45 

40 

35 

30. 

^=11.2 12.0 12.93 

14.0 

15.3 

16.8 

18.7 

21.0 

24*7 

28.0 

For ocean waves the 

following are 

given by I>r, 

, G. Schott 


Moderate wind, Beaufort 

Scale 

5 

= 28 

m.p.hr 

L. - 33 h. 


Strong wind, " 

It 

6 to 7 

= 35 

It n n 

L s 20 h. 


Storm " 

n 

9 

= 56 

n ft tt 

L = 17 h. 



127 


















The ocean waves are thus relatively longer than waves 
on the Great Lakes,as might be expected. 


b> Height of wave ab o ve stil l water le vel . Since the 
crest of a wave is above the still water~Tevel and the trough 
is below that level it is sometimes erroneously assumed that 
the still water level represents the mean of the two. How¬ 
ever, many observations made by various observers show the 
crest to be about Eh/s above the still water level in deep 
water, and raising still higher in shallow water^ a fact which 
must not be disregarded when examining the safety of a struc¬ 
ture. 

According to many observations made by Capt. Gaillard 
on lake Superior, the height of a wave crest above still 
water level before the wave breaks, is given by the formula: 


a. » il + Jl for deep wafer Wavei, wiiVi d > l.84-h 

2 I 

a = ^ for shallow wafer waves, with d < J.S^h J 58-B. 

For deep water ocean waves, Rankine gives the formula 

a = ^ 4-0.785^..58-C. 

The term still water level as used hc're, represents a 
line for which the sectional area of the wave ridge is equal 
to that of the wave hollow. 


c. Depth of wa ter in w hich a wave brea ^. While d#ep 
water waves may brealt Hue to the action oT~increasing wind 
and other causes, yet this is invariably the case when a 
wave reaches water of insufficient depth. 

A knowledge of this minimum depth is necessary in 
determining the maximum wave which may be expected to arrive 
at a certain structure located in shallow water. 

Undoubtedly the roughness of the bottom has much to do 
with this subject and it would seem that the depth in which 
a wave breaks in much less for a rough bottom than for a 
smooth sandy bottom. 

For 55 observations at the Duluth Canal by Capt. Gaillard, 
for waves from 7 to 13 ft. in height and a sandy*bottom sloping 
about 1:40, the waves broke in a depth equal to 1.72 h. Similar 
observations near Presque Isle Pt. and near Grand Marais, Lake 
Superior, with rough bottom and waves from 6 to 9 ft. high, 
the waves broke in a depth equal to 1.3 h. ’ 

For ocean waves at St. Augustine, with strong wind blow¬ 
ing in the direction of wave travel, the saves broke in a 
depth of 1.25 h. 


128 








Henoe, calling, d, the depth in which a wave breaks, 
and the height of a deep water wave from troiigh to crest, 
the following average relations may be accepted for storm 
waves with wind in the direction of the wave travel. 


Lorkc Super’ior, sondy bottom , wind m.,d = l.84h 

•• *• .. •• .. V=2G »n.,d,=. 1.4-2.h 

•' •• •• breeze,d,si 1.34-h 

•' •• rou^jh bottom , wind Vs 30 m., d|= l.3oh 

Ocean sandy bottom , wind V= *n., d, = 1.25h 


V 58-D. 


^t. 59 . Dynamic Pr opert ie s of Wa ves. The previous 
article dealt wilh wave "dimensions, and IKe relations exist¬ 
ing between these and the actuating force. The present 
article will be devoted to the dynamic properties including 
the velocity of propagation, energy of a wave, wave pressure, 
and height of a completely obstructed wave. 


a. Vel ocity o f p ro paga tion. According to Rankine for 
long waves or waves in”water that is very shallow coaparod 
with the wave length, the velocity is given by the formula: 

Vz^]j^(d -h ih) = -59-A. 

This formula according to Gapt. Gaillard, invariably 
gives results in excess of observed velocities. 

The theoretical velocity for deep water waves with 
d > L/z is: 

= 2 - 26 ( 1 :.-. . 59.B. 

For shallow water waves with d < , the formula becomes 


Vz^Z.ZGcfZ when ^ + Z _ 

where c has the following values depending on <Vl 


59-C. 


= 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 

c = 0.552 0.746 0.858 0.922 0.958 0.977 0.988 0.994 


The wave lengthL and velocity v are materially altered 
when the depth gradually decreases toward a shoal or shore, 
and Gapt. Galllard proposed an empirio formula based on a 
number of observations made at Duluth Canal, where the waves 
entered the canal at 26 ft. depth, while the portion of wave 


129 











outside the canal entered shallow water, gradually shoaling 
to 3.3 ft. The formula is; 




59-D. 


in which v is the velocity of the wave approaching from a 
depth d which should not be greater than L/z,and 14 is the 
reduced velocity when the depth has shoaled to the lesser 
depth dj . A wave thus retarded may have a reduced height 


o-f h ^2 as per 59-D. 


In applying the formulas 59-D it is well to choose a 
deep water depth about L /5 instead of L/z because the 
dimensions of a deep water wave ere not appreciably altered 
until after the shoaling has become effective. 


b. The energy of a wave. The theoretical energy for a 
deep water wave of length L feet, height h feet, and breadth 
of one foot, for fresh water, is given by 


E = 7.8UW"[,-4.93s(^)^J«.lbs.. 

For salt water the coefficient 7.8 becomes 0.0. 

For shallow water waves the energy decreases from 
2 to 10^ below the value given by Eq. 59-E and the appropriate 
formula becomes rather complicated. No particular use is 
made of this formula so the matter will not be investigated 
further. 


c. Wave Pressure . Having enumerated the various steps 
in arriving at the probable dimensions of storm waves for a 
given locality, the important question confronting the 
designing engineer, is to appraise with some reasonable degree 
of accuracy, the force which such a wave develops in striking 
a proposed structure. 

The most satisfactory way of attacking this problem is 
by direct measurement of the force expended by impinging 
waves on a solid structure carrying a series of selfregister¬ 
ing dynamometers. 

Dynamometer measurements were carried on for many years 
by Thomas Stevenson at Skerryvore Rocks in the Atlantic Ocean 
in 1843-4, also at Dunbar Harbor in 1858. 

Several attempts to measure wave pressures on the Great 
Lakes were made at various times, but without gaining much 
information of value. The experiments made, in 1901-3, by 
Capt. Gaillard on Lake Superior,stand as the most valuable 
contribution to this subject. 


130 







Before mentioning any experimental re8ults_,the following 
conclusions arrived at by Capt. Gaillard as a result of his 
careful and painstaking work are here quoted* 

1* That the impact of a wave does not at all resemble 
that of a solid body. 

2. That the pressures indicated by the types of 
dynamometers heretofore used are due to dynamic action only. 

3. That these pressures apparently conform to the 
hydrodynamic laws governing the action of a current flowing 
normally against a submerged plane. 

4. That a mass of water in air projected with a 
certain velocity against a plane surface can produce no 
greater pressure than would be caused by the steady flow 
against this surface of a Jet of equal cross-section having 
the same velocity and striking at the same angle. 

Prom the above statements, it apparently follows 
by inference. 

5. That a mass of water projected against a submerged 
plane surface of considerably smaller area than the cross- 
section of the mass,can produce no greater pressure than 
would be caused by the steady normal flow at the same velocity 
of a current against a submerged plane surface of equal area 
and similar to the first. 

6. That as the most destructive waves act for an 
appreciable period, the pressures which they exert can 
properly be measured by suitably constructed dynamometers. 

Observations made with spring dynamometers on the 
outer end of the South pier, Duluth Canal, Lake Superior, 
by Capt. Gaillard, are given as representing the best typical 
wave measurements available. See Table 60-A. 

The formula for pressure on a submerged plate according 
to Dubuat is 

lbs. per 

in which v is the velocity of forward motion of the plate. 

The total velocity of the striking wave consists of the 
combined velocity of propagation v and the maximum orbital 
velocity u;, of a wave particle, so that the above formula 

should be modified to represent the maximum value 


p - k lbs.per sq.ft-59-P. 

where k is an empiric coefficient evaluated from Capt. Gaillard*s 
observations for the Great Lakes as 1.30 to 1.71 for winds 
from 30 to 70 miles per hour. For ocean storm waves k may be 
taken as 1.8. For sweet water w= 62.4 lbs. cu. ft. and for 


131 




salt water w = 64,4 lbs. while the acceleration due to 
gravity is g = 32#2 sec. ft. 

According to Eq. 59-C the velocity of wave propagation 
for waves formed in any depth of water is : 

Vs 2.Z6 C \fL , where C=Vi2 with the ratio o-f the 

» ’ I <»5 a* 

semi-axes of the surface orbits as previously given. 

The maximum orbital velocity of a wave particle is 

m , where m = =the semi-major axis of the 

elliptical orbit and p is the ratio of the axes of the 
elliptical orbits. 

The numerical values of c and p , given in Table 69-A, 
render these expressions applicable to the solution of 
problems. 


Table 69-A - Values of c and g for values of do/L . 


U L ''L/ 

c 

? 

ii 

0 

0.06 

0.662 

3.266 

1.814 

0.10 

0.746 

1.796 

1.340 

0.16 

0.868 

1.368 

1.165 

0.20 

0.922 

1.177 

1.086 

0.26 

0.968 

1.090 

1.044 

0.30 

0.977 

1.047 

1.023 

0.35 

0.988 

1.026 

1.013 

0.40 

0.994 

1.013 

1.007 

0.46 

0.997 

1.007 

1.004 

0.60 

0.998 

1.004 

1.002 


Introducing the numerical values of IT and g,the above 
velocity formulas become: 

Us 2.ZG C^TT and =, m^=7-11 > 

and with values of h, w and g substituted into Eq. 59-P, 
the maximum unit wave pressure becomes: 

ps k ^ lbs. per st^.fh.- 

p=.l.7l (^\f ■*'‘^ 0 ) fresh water storm waves. 

p = 1-80 for salt wafer storm waves, 

where = 2-26 +■ 7.11 - 



132 















with values of c and as given In Table 59-A, The values 
driven for k are maximum for 75 mile wind and may become as 
low as 1.3 for 30 mile wind. 

When do/L ? then c = jw. = i.oo , which is the 

condition for deep water waves. 

Having determined the maximum pressure exerted by a 
striking wave, it is necessary to know the elevation above 
still water level at which this maximum occurs, also the 
height to which a completely obstructed wave acts. 

The maximum wave pressure occurs at a height h, above 
still water level, the theoretical value of which is: 


h,= o .78 5 h^, while 


gives better values--59-H. 


When a wave is completely obstructed by a vertical 
surface of sufficient height, the wave crest is raised to a 
height equal to above the still water level, and at 

this height the pressure becomes zero. 

With this data it is now possible to design the entire 
wave pressure area and thus evaluate the total force to be 
resisted by a proposed structure. 


d. Oblique waves striking a breakwater produce less 
pressure than when striking normally. Using the relations 
indicated in Pig. 59-A, the wave pressure p, Eq. 59-P, is 
reduced to a value 


p 


59-H, 


breakwater. 



Observations made by the 
Author at Toronto, Ont., in 
1915-16 gave the following 
result: 

p = 14-6 0 lbs.per 

p = 820 lbs. per sq.-ft. 

oC =1 50®, si <?C =. 0.587 

By formula p = )4-6Ox0.587=857lbs. 


Fi§.59-A. 


133 






Art. 60. wave pre ssure observtatioub Lfake Su per ior made 
in 19T)l^"by Capt. D. D. Jaillard, Table 60-A, are driven as 
representiiu^ the best available data and will serve as a 
pattern oi wave pressure curves. It is regrettable that 
no accurate wind velocities were recorded although it v/as 
stated that moderate wind, from £0 to ^5 miles, produced 
these waves at Duluth. The pier carrying the dynamometers 
was not of sufficient height to completely obstruct the larger 
waves, so that for waves over 13 ft. high the recorded pres¬ 
sures are for partially obstructed waves. 


Table 60-A Wave Pressure Measurements, Duluth Canal, L.Superior. 



Elev. 

still 

water 

O b s e rvc d 

Max. wove dimensions 
Depth d=25-ft. 1 

-» - -j 

Max. DynamometerReadin^s. 
Pounds per ft. 

^ -- 

Computed 

Date 

Height 

Length 

Velocity 

Elevationsof Dynamometers 



P = 


level 

h 

L 

TT 

4 . 0 . 07 ft 

-h 3.74 ft. 

+- 7 . 01 ft. 

L 

x 

r-r^ 



ft. 

tt. 

ft. 

Sec.ft. 





sec.ft. 

lbs.S 4 -ft. 

1901 
July ^4 

t- ‘ • 

12 

ISO 

24.2 

250 

1 1 50 

I 0 30 

0.180 

7.8 

1330 

Auq. 9 

^ 1.9 

12 

130 

24.2 

370 

1 190 

780 

1 

0.210 

8.0 

1348 

Oct. 9 

1.9 

13 

150 

29.6 

0 

1 fc 1 5 

1260 

O.lgo 

8.4 

1876 

NoV.22 

1.4 

14 

ISO 

27.2 

0 

160 5 

160 5 

0.184 

90 

1 705 

Sep. Z4 

4- 1.9 

IS 

250 

33.2 

1630 

225S 

2 0 50 

O.IIO 

9.3 

2344 

1902 





-1-7.04 ft. 

-H2.57ft. 

4- la.lftft. 




Oct. 25 

+ 1.7 


200 

SO.O 

1755 

1335 

0 

0.138 

9.5 

2026 

Dec .20 

4- 1.7 

I& 

210 

31.0 

1 700 

14-30 

S15 

0.131 

9.4 

2 1 20 

Nov. 12 

^ 1.7 

IS 

250 

■32.0 

2370 

2195 

13 70 

O.IIO 

lO.S 

2344 


*■ Values o-f ^ ore taken from Table S9 A. 


The coefficient 1.3,in the formula p = l-3(^u+, was 
chosen because none of the above waves were caused by extreme 
storms,while the possible maximum coefficient 1.71 Ariven for 
Kq. 59-F is more probable for winds from 70 to 75 miles per 
hour w'ith shorter waves. 

In order to arrive at the total wave pressure to be 
resisted by a sea wall or breakwater, it becomes necessary to 
design the complete wave pressure curve from data previously 
calculated for a given wind velocity and fetch. 


134 












































We thus obtain the wave dimensions h, L , a and finally 
determine xr , and p . It still remains to find the total 
wave pressure for a completely or partially obstructed wave 
of these dimensions, and this must be approximated by 
patterning after some typical set of observations like those 
shown on Pig, 60-A. 

By selecting the maximum pressure values for a 16 ft, 
wave from Table 60-A, into a composite pressure curve, the 
section shown in Pig, 60-A was obtained. The orest of this 
wave was 10,8 ft, above the still water level, and this crest 
was raised to a haight of 18,3 ft, for the partially obstructed 
wave with the trough at 6,18 ft, below the still water level. 
Had the wave been completely obstructed, its creat would 
have reached a height za. = 21,6 ft. The straight line 
AC is drawn to represent the maximum pressure of the completely 
obstructed wave with the maximum pressxire at h, =o.izh 
above the still water level. The pressure curve is completed 
by drawing the curved line CB. The total wave pressure for 
the oompletely obstructed wave is represented by the triangu¬ 
lar area ABC = s 35 , ieo ib^., acting at the center of gravity 
of the pressure area which is 5,8 ft, above still water level. 

For the partially obstructed wave as observed, the total 
wave pressure figures 33,340 lbs., acting at 5,42 ft, above 
the still water level, 

A composite wave pressure curve as above described, 
represents the plotting of the maximum pressures recorded, 
for the duration of a storm, on the several dynamometers 
located at different elevations. It is evident that this 
record is not produced by any single wave, but that waves 
of various sizes contribute to produce the maximum pressures 
recorded. Hence, the total pressure area of such a curve 
is undoubtedly greater than the actual and thus includes a 
certain factor of safety. 


"35 



A COMPOSITE WAVE PRESSURE CURVE BASED ON OBSERVATIONS 

FOR A I6FT. WAVE IN LAKE SUPERIOR AT DULUTH. 

Observations made in I90I-2 by Capt. D. D.G-aillard,Corps of En^rs.^U.S.A. 



136 































meve pressure observ atio nb, Lak e Ontario, made by 
the author lor the Toronto Harbor Commissioners at Toronto, 
during a storm on llov. 19, 191b. The storm commenced at 
b P.M. on Nov. 18, with east wind, and reached a maximum 
of 42 miles per hour about 7 A.M. on Nov. 19. 

The observations were made at 10 A.W. with the wind 
at 30 miles after which the wind veered to the S.E. and 
died at 2 P.I!. after swinpin^? to the south. 

The pressures were recorded on spring dynamometers 
mounted on a solid crib situated about 700 ft. off shore in 
about 8 ft. of water, ^t a distance of about 1200 ft. from 
the crib the water deepened rapidly to about 40 ft. In 
front of the crib, three piles were driven at about IbO ft. 
centers forming an equilateral triangle and carrying g&ge 
boards for observing wave height, also direction and velocity 
of wave crests. The results of these observations are shown 
on ij'igSi 60-B and C. 

The deep water wave dimensions Fig. 60-D were estimated 
for a 30 mile wind 'and a fetch of 100 miles, representing 
something like an average condition prior to reaching the 
maximum. The still water level was at El. 245.0 out in deep 
water and began to raise when the depth was reduced to 17 ft. 
where the waves broke, and finally reached the Elev. 247.2 
on the beach. 

The reduced wave after entering shallow water, and 
producing the record shown by Pig. 60-C, had a height of 5.5 ft.^ 
length 110 ft., and velocity 19.8 ft. per sec. These dimen¬ 
sions and pressures were checked by applying the formulas 
previously given to show the agreement between computed and 
observed values. 

It is thus seen that for a given condition of wind, fetch 
and depth in which a proposed break^water is to be built, 
the wave pressure curve may be estimated with sufficient 
accuracy to test the stability of the proposed structure. 


'•37 




WAVE OBSERVATIONS ON LAKE ONTARIO AT TORONTO, 

STORM N0VJ9,19t5, USING SPRING DYNAMOMETERS. 

By David Molitor,C.E. 



138 







































wind Roses . showint^ maximum v/ind velocii'ies far the 8 Card\na\ poinVs, 
and the dates on wKich they occurred. Data obtained from U. S. vVea+Her Bureau 
Officials in charc^e of the Several Stations. 



S 

Oswego, H.Y. i87l-l93l. 




Toronto,Ont. 1911-1916. Buffalo,N.Y. I9»^l—1929 


N 



H 



DetroltMich. 1913-1931. 


N N 




S 

Port Huron, 1874 - 1930 . 



All the obove wind velocities are based on 3-Cup anemometer voiues. 

Fifes. 60 E. 


139 

























The following data reKsrdlng weights ahd friction will 
be useful in designing breakwater oribs. 

Weights in pounds per cubio foot« 

Material Weight in Air Weight in Wat er 

Granite blocks - 170 - 107.6 

Limestone blocks_166--103.6 

Limestone rip rap - 40?^ Voids.- 100_62.6 

Concrete, plain_148_06.6 

Conori^te, reinforced-162_89.6 

Timber, green as shipped_36__uplift * 27.6 

Timber, wet, incl. bolts_60 _ .uplift - 12.6 

Ayerage timber cribs, 

filled with limestone-92-60 


Coefficients of friction 

All surfaces of masonry or brickwork in contact 0.66 to 0.70 


Stone or brickwork on moist unctuous olay_0.3 

Stone on bed rook, dry_0.7 

Concrete blocks on well wetted concrete floor_0.7 

Concrete blocks on rubble base, submerged_0.66 

Rubble filled cribs on rubble mound, submerged_0^9 

Rubble filled cribs on bed rook, submerged_0.6 


140 


















Art* 61 Stability of Brealc water Crib s subjected t o 
Ware Force. The previous articles present the several 
steps in evolving the wave pressure curve from a given fetch 
and wind velocity * or in other words, of evaluating the 
wave effect produced by a certain causes The ultimate aim 
is however, to design a brealnvater crib of adequate dimensions 
to safely withstand the wave pressure to which it may be ex¬ 
posed during the most severe storms* 

The complete solution of such problems will now be 
undertaken on the assumption that the exact location for a 
proposed structure and the maximum wind for the locality are 
definitely known* The examples chosen are the Breakwater 
at Harbor Beach, Mich*, and Magann^s Pier at Toronto, Ont*, 
both of which structures have withstood many severe storms 
during the past 30 years without showing any marked structural 
weakness* Hence, if the method of analysis here given leads 
to findings in conformity with the actual behavior of these 
structures, then the speculative features involved in the 
methed will have been reduced to a negligible quantity* 

Problem 61-A * Stability investigation of Harbor Beach 
Brea^ater * See Pig* 61-A and the computations which follow* 
The fetch, scaled from a map of Lake Huron, is found to be 
132 miles north or 92 miles N*E. The latter fetch is about 
normal to the breakwater* The maximum wind velocity from 
the N. is taken as 60 miles per hour* Since waves usually 
swing more or less normally to the shore as the water becomes 
shallower, the larger wave is assumed to arrive normally 
in front of the wall* Eq*57-A then gives the probable 
height of wave as 15*13 ft* and Eq. 56-A gives its length 
as £1£ ft* Since the water in front of the breakwater has 
a depth of 28 ft* this wave suffers no appreciable reduction 
due to shoaling effect* Hence the problem is comparatively 
simple and requires no further description aside from that 
given* In estimating the factor of safety no account was 
taken of the weight of water on top of the superstructure* 


141 












BREAKWATER AT HARBOR BEACH,MICH. 

STABtLITY INVESTIGATION. 


Harbor 

'4. 



Ma'.s concrete 


Pr: 27.424 lbs 


Stonefill 


Bloclr 


block 


Block 


Concr. 










|aS3BgBSS5g^toS^ 




Bssssssasss 


jmaa&aaerieiasgg 


wave c're&y 


-rl6.>5 


yyaYC t ^ j i > -fro. 

i"£____ / jaiLfitl _ A_+Js 

r ^ T \ 


ixle! 


5-0- J 


Level. 


_ 1 _ 


-6.4-a 


Safety a^inst overturn. 


SsriSO K 18.82 r>8 

274)4-* 31. S 


Safety a^inst sliding 


9S| so X0.5 _ I 74 . 
27424 

The mo*, unit pressure 
atihe Weelof the wall is 


_ 9SI SO H 2 _6470n 
3* 9.9 

Bottom, 


3S-0 


Fi§.6l-A. 


The feteb D =.I 32 m. North; Angle c3C = 9o“; Wind V=.6o m.p.hr. Hence the deep Water 

wave by E<^.57A has b = o. I7\|VD =15.13 ft. and by Eq.SSA find L = 14.0 h = 212 ft.j d=.28ft. 

The height of this wave above Stillwater level by Ea. S8B is a= h . g.esft.. and h-a=.6.48ft. 

/ ^ ^ TT ' 

This wave will break when the depth becomes d^=. I.84h = 27 . S4ft., hence the wave will 
strike the breakwater before breakint^, ond will suffer no material reduction due to 
shallow woter. 

Now -find +-2 0.142, for which Table S9A <jivcs C=0.829 

and ^= 1 . 21 . Then E(|,.S9G (jives u■-^ V, = 2.26 efU -t-riiyk^ = 36.2 Sec.ft.anel the maximum 
unit wave pressure becomes p= 17 (v 4 -u;)^ = 22^0 Ibs.per Sg.ft. actin<j of h=o.i2h=l.82ft. 
above Stillwater level. Then make ^ — 0.72 p = l600 lbs.S(j.ft. 

This data determines the complete wave pressure curve as shown above, 
which requires slight modificotion for the portion of wove which is not obstructed. 

The total effective wove pressure is thus found as P= 27,424 Ibs.per lin-ft. Qctin(j 
31.3 ff. obove the bottom. The weight of the crib, allowing for buoyancy, is calculated 
QS G= 95,1 SO lbs. per lin. ft., which Combined with P gives the resultant R octing on 
the base and producing a pressure of 9 5^1 50x2 6470 lbs.per sg.ft.at the heel. 

No account was taken of the weight of the wafer on top of the Superstrucf ure. 


142 






























































































































Referring agftin to Fig, 61-A, It will be noticed that the 
wave force P was combined directly with the weight G in test¬ 
ing the safety of the structure, without regard to the shape 
of the concrete deck or superstructure. This would be correct 
in testing the stability of the crib, provided that the entire 
wave impact is imparted simultaneoucly to all parts of the 
structure. 

However, since the superstructure does not attain its 
full height at the face of the crib, but slopes gradually 
for a distance of 18 ft, and then presents another surface 
to resist the upper portion of the wave, it must be apparent 
that the full wave force is not expenia^ simultaneously, nor 
with the pressure assumed in the analysis. The wave could 
not be built up to its full height until after it had advanced 
to the last obstruction a second after striking the first. 

In other words, the total wave energy is transferred to the 
crib during a greater period of time than was assumed in 
designing the wave pressure curve, and hence the shook is in 
reality less severe than the calculation shows. 

This example illustrates how the wave pressure may be 
rendered less severe simply by adopting a superstructure 
design which does not obstruct the wave in its entirety on a 
vertical surface, by allowing the wave to build up gradually. 
However,this may develop some difficulties in the design 
of the concrete superstructure end no doubt will cause a 
larger proportion of the wave to Jump over the wall. The 
back wash may also be retarded to such an extent as to meet 
the next oncoming wave at an unfavorable time. The wave 
force striking the high center wall will subject the latter 
to a moment of about 9600 ft, lbs, which might cause the 
mass concrete deck to break unless steel reinforcement ie 
employed, 

Considering the ample safety of the crib itself, it 
would appear that the high mass concrete wall might better 
have been placed over the lake front thus affording greater 
security for the concrete face blocks, besides adding very 
materially to the strength of the concrete deck where it 
receives the worst punishment. 

In designing a breakwater superstructure, much depends 
on the purpose which it is to serve. It may be necessary 
to completely obstruct oncoming waves, especially if ships 
are Intended to moore on the harbor side. In other cases it 
may not matter how much water ie thrown over the wall so 
long as the wave force is rendered harmless. 

The superstructure of the Harbor Beach breakwater, if 
intended to protect ships while tied on the harbor side of 
the breakwater, might have been designed as indicated by 
a dotted outline. Fig. 61-A, so as to obstruct more completely 


143 


the oncoming waves, at the same time increasing the resisting 
moment of the structure and adding strength to the deck. 

The use of some reinforcing bars in connection with these 
superstructure designs is very desirable. 

It is also necessary to provide breathing openings 
through the concrete deck to allow relief for air pressure 
which is formed in the body of the crib. Wave pressure will 
also set up hydrostatic pressure on the inside of cribs, thus 
exerting a bursting effect on all parts of the structure which 
must not be overlooked. 

Problem 61~B. Stability Investigation of Magann’s Pier , 

Fig. 61-B. This pier is located between Jameson St. and 
Dowling Ave., Toronto, Ontario, in 9 ft. of water, deepening 
to 11 ft. at 60 ft. out. The worst conditions exist for a 
S.E. wind with a fetch D = 30 miles, making an angle oc = 68° 
with the lake face of the pier. A wind velocity of 56 miles 
per hour was taken as a possible maximum. 

With this data, the deep water wave dimensions may be 
estimated by Eqs. 67-A and 68-A, with a velocity of propaga¬ 
tion u, figured by Eq. 69-C for a depth cl - L/5 . This 

depth is arbitrarily assumed because a deep water wave may 
continue into much shallower water before the shoaling 
effect becomes appreciable. 

The important shoaling effect is produced when the wave 
advances from the depth d =L/ 5 =s 2 i ft., to that in front 
of the pier which is only 9 ft. The reduction in velocity 
and height, due to this shoaling, is evaluated by Eqs. 69-D 
and the pressure exerted by the wave so reduced is found from 
Eq. 69-G. Kow since this wave does not strike the pier normally, 
a further pressure reduction is made as per Eq. 59-H, but this 
does not alter the wave height just found. The pressure area 
may now be drawn for a completely obstructed wave, reaching 
the height 2a=i 5.6-2 ft. 

In this problem, the pier is not sufficiently high to 
completely obstruct the wave, thus necessitating a reduction 
in the pressure area for the portion of wave which is not 
obstructed. The final pressure P is represented by the area 
of the pressure curve between the wave trough and the top of 
the pier. The detail calculations are given following the 
drawing, and show a stability against overturning and sliding 
on the base quite in conformity with the behavior of this 
structure during its 30 years of existanoe. 


144 







shallow wo+er and ^he wave suffers a reduction in site and vclocit'y.Accordin<j+o Ec^.SSO, 
this wavc will break when the depth cs reduced to d,= t.84-h= ll.7ft. Just before stnk'm^thepier^ 
the depth is further reduced to d^a ^.O ft;cau sin<^ the initial velocity V to retard to a value 
according to 59 D. Thus t^ = o-9 sec.ft., ond the wave So retarded is reduced 

to a hei iht h^.hf|7»«., ^^2 —9.99 ft. with a length L= i5.ihj^=S8.7ft.The height otthis 
wure above still water level by Eia.SSB is hz ^ z il* = z ^i53^ = 7"8lft..andh-3 = l-68ft. 

Also f-ind 0 . 2 GI,for which Table 59A gives )u.. i.o9. Then by Eg.SS^r 

find 1^4. iT^ ar 19.5s9- 7.11» 1.09*s ZO.S Sec.ft. The maximum unit pressure then becomes 
p = 1.5 1.5* Z^'ss 59’l*i'Zs.per sg.ft. octing at hj= 0 .izh^= 0 .S 9 ft. above still water level. 

This pressure resolved normally to the p'ler aspcrEg.59H becomes = p sin^<XaS46x0^896=962 lbs. 

From the above data construct the final pressure curve as above indicated to obtain 
the total pressure P =s IS 80 lbs.per lin.ft.of pier. The weight of the crib,allowing for buoyancy, 
is G-=6180 Ibs.per ft., which combined with P gives the resultant R acting on the base AB. 


145 






































Art> 62 . Conoludlnp Ramark:a« In closing this subject 
the Author wishes to emphasize that stability j.nve8tigation8 
of this character, or designs based on similar calculations, 
must be regarded as more or less speculative , yet the 
methods here presented for the first time, were developed 
after much thought and study and are considered as accurate 
as the subject warrants. 

The same method of analysis was applied to existing 
hrealnwater structures at Presque Isle, Ontonagon, Grand Marais, 
Agate Bay, Portage Lake and Marquette, all on Lake Superior. 

The Buffalo breakwater comprising several designs with varied 
exposures was also analysed. In each case the results of the 
analysis agreed elosely with the behavior exhibited by these 
structures during many years of service. 

The western breakwater at Toronto, built in 1915-16, 
according to designs which were criticised at the time as 
being structurally inadequate, showed a factor of safety of 
about 4 against both overturning and sliding, according to 
an analysis made in 1915 by the author. This breakwater, 
covering a length of about 3-l/£ miles,has weathered all 
storms for the past 15 years without manifesting the slightest 
signs of weakness, again demonstrating the reliability of 
results obtainable with the method here described. 


146 




ELEVATED STEEL CYLINDIilCAL TANKS 


Art* _ Tank Bottom a Surface of devolution. In this general case the bottom 

plate is curved in two directions with radii^and n as shown in Fig.l. The radius 
§ is the radius of curvature of the plate at any point P for a vertical section. 
The radius nsg^^is the normal to the surface at P, and is the radius of a meridian 
circle through P with its center on the axis of revolution at N. For a conical 
bottom,^ =infinity. 


The principal stresses H and S, per unit length and width of plate at 
any point P, are related to the normal unit pressure p acting at P, and this 
fimdamental relation will now bo determined. 



For a vinit pressxire p on the plate, the total pressure on the small 
area d^will be pd* H and S are the stresses per \anit edge dimension of plate 
in the two directions. 

The tension along the edge of the plate area d^will be Sd in the vert¬ 
ical meridional piano, and Hd in the horizontal direction,as shown in Fig.l-b., 
where S and H are stresses per xmit width in the two directions. 

From Fie«l-c, by similar triangles,^z=Sd; z , giving z= , which 
is the con 5 )onent of Sd in the direction of the normal pressure p at tra center 
of the plate area d^. 

A similar value z'may be foimd for the forces Hd , acting horizontally, 
as z'= Hd^, whore nsP K=^ . 

2 n Stn<k. 

Now since the sxim 2(z-hs^), must equal the total pressure pd^, then 

Sd^Hd^-Pd^ or p = & or H = n( p-S) --- C* ) 

Y n'' S ? 

which is a fundamental relationship between the unit pressure p at any point 
of the curved plate, and the stresses H and S , acting respectively in a horiz¬ 
ontal and a vortical direction or plane. 


147 
















Now it is always possible to evaluate the stress S per luiit width of 
plate, in terms of the water load G, acting on the tank bottom and within the 
cylindrical volume of radius r. Then with S and p known far any point P, the 
horizontal stress at this point is given by Eq*l ,as H=: n(p- S). 

Thus the vertical component of S Is S slncC per unit of length of plate* 
and acts around the entire circumference PTTr of the small circle with radius r» 
The total vertical shear 2TfrS sinocaroxmd the circumference STTr, must then be 
et|ual to the vortical water load G contained within the cylinder of radius r* 

Hence, Gr^S'TTrg sincx . or 3 = G - ----^ 2 ) 

21T r sinoC 

Equations 1 and 2 thus afford a solution for the two principal stresses 
at any point P in the bottom plate of a tank, shaped as a stirface of revolution. 
When the bottom is conical, then ^ is infinite and Eq. 1 gives Hsnp, which is 
independent of S. 


Hemispherical Bottom Tanks . The volume of liquid of depth y, above the 
tank bottom is 

iry^(B— y/3) 


The volume of the cylinder of radius r and height h—y is 
Tt r^(h—y) = 'TT (2R—y) y(h—y) . 

Henve, the total weight G resting on the bottom within the circle of 
radius r, due to water weighing w lbs. per cu, ft., becomes : 



Noting tnat r sinx = (2S—y) y , Eq(2 ) gives the value of S as ; 

H TJ 


S- 

[(2R-y)hy-RyV-|-y^] 

1 _ wR| 

rh-ypR-2y)-] 



2' 

L 3UR-y/J 


The value of S from Eq.4 , substituted into Eq.l , and noting that 
n = (|= H for the hemispherical bottom, then 

H = pR—3 = wR(h--y) —3 = ^ j^h—y |9R~-4s^j _ 

The stresses H and S become minimum when y=R, giving 


148 



























for all points at the levol of the center 0, or distant R above the bottom of 
the tank. 


So long as the tank is full and h— R/s does not become negative, then S 
remains a tensile stress. However, to avoid compressive values fpr H, the value 
of h must be eqvial to or greater than 5R/3 • This would require the tank volume, 
including the cylindrical portion, to be at least or 


R= 0.62'^ Volume of tank. -(7 ) 


When the water level recedes to a depth h , less than 5R/3 , then , accord¬ 
ing to Eq.6 , H becomes negative or conpressive, while just above the hemisphere 
H * R(h--R)w. which is tensile. This stress reversal at the top of the hemisphere 
should be safeguarded by a horizontal girder provided at this level. 


V/hen the water just fills the hemisphere with level at 0 , then h=: y=: R, 
and ^ —h. at the water level. 

“ 3 ” 

When the water level drops below 0 to a depth y , then 



When there is a vertical stand pipe connected with the tank bottom, thus 
relieving tlie tank of it' water load, then the above Eq.2 and all subsequent 
formulas will require modification to allow for that portion of the weight G 
which is then carried on a separate foundation imder the stand pir^e. 

A rt. _ Tank Supports . V/hen a tank is htmg from a circular wall, the matter of 

peripheral support is iq[uite sinple and resolves itself into providing a suitable 
bearing plate to transfer the uniform tank weight g to the masonry. 


However, when a tank is supported on a tower, the weight must be carried 
by a balcony girder to the tower legs and thence to footings in the ground. These 
structural elements will also be called upon to carry the lateral wind pressure 
acting on the tank and the tower legs. 


The balcony girder is considered fixed at the colvimn supports A, has a 
radius E, and subtends a center angle 2oc,Fig.3 . 

C^(t-coeeC) Lot G - the total weight of tank and contents 

carried on n tower legs. 

=load per lin. ft. of girder. 

2irs 

P K 2gocR = — •=. the load carried by one column, 
u 

Thon the bending moment at the center of the 
beam, considering both ends fixed, is approximately 

2R 8inac)\^i^gH^-(9 ) 

The bonding moment at the fixed ond A is 
approximately 

^ il^CQ8QC-gR^(l—cos«k.)- (JO) 

Also the torsional npnent at the fixed end A is approxiiiiatoly 

gR^(«>c- sinoc)- M^sinoC -(II ) 



149 






















For tank towers, there aro relatively few cases to be considered and the 
following T'ablea will avoid the necessity of evaliiating the above formulas. 


Table ocFunctions for Balcony floarns. 


No.Posts 

oC 

OC 

sinoc 

008 oC 

1— C08(X 

oE— sinoC 

in tower 

Degrees 

Arc 





4 

45 

0.7854 

0.7070 

0.7070 

0.2930 

0.0784 

6 

30 

o.5see 

o.deeo 

0.8660 

0.1340 

0.0236 

8 

22.5 

0.3928 

0.3827 

0.9239 

0.0760 

0.0101 

12 

15 

0.2619 

0.2588 

0.9659 

0. 0340 

0.003* 


Table Bending Moments and and Torsional Moments • 


NO.Posts 
in tower 

Center mom. 
g sin^oc gl? 

End moment Ma=- 
U^cosc3C-g R*^l— cosoc) 

Torsional mom. T^,- 
g r( ctC —sinoc)^-sinoc 

4 

0.0833 gR^ 

0.2340 gK^ 

0.0195 gR^ 

6 

0.0417 

0.0980 gR^ 

0.0027 gR^ 

8 

0.0244 gR* 

0.0536 

0.0008 

12 

0.0112 gR2 

0.0813 gS^ 

0.0002 gR^ 


The required section modulus 3 for the balcony beam is given by th<» 
approximate formula: 

3 _ Mg 1^1 -»-2.9 ^ Ta.j inch pounds, — — --(12 ) 

where f is the allowable unit stress in pounds per square ino-h. The formula 
gives good results for ‘^-c M • 


Also, the combined stress f at A, due to bending and torsion, is given 
by the approximate formula 




Ix 


I+ 


for M and T^ in inch^poundSf- 


('3 ) 


where and ly are the principal moments of Inertia of the beam, while e and e' 


are distances from the c. g. to the extreme edges of the section for bending 
and torsion respectively., ^ | - 


The connection between two adjacent balcony beams to the top of a 
column, will transmit a bending moment to the oolxunn equal to 2T , and due 
to elastic deformations in the several balcony beams, will tend to bulge the 
cylindrical tank plates out between column supports. Hence, the. necessity for 
a heavy, stiffening element at the top of tho‘ cylinder to maintain the circular 
shape. 


The colxnnn load V will c'onsist of the dead load G/n=P, combined with 
the transferred wind load. Tha vertical coa^jonent of the transferred wind load 
is downward on the leeward columns and upward on the windward columns, and is 
maximum when tho direction of the wind is taken parallel to a long diameter of 
the tower. Thus, for a wind moment M, Fi^.4, 

V= P ± R 
I 


150 






































where 


I is the nornent of inertia of the n\unbor of columns, 'fhus: 



y 




R/z' 

m 









y 


for 4 

cols. 

II 

and V= P i 

LI 

2 R 

6 

*« 

31?* 

P ± 

M 

3 R 

8 

ri 

4R^ 

P ± 

M 

4R 

12 

ft 

6E* 

P± 

M 

6 H 


Plan of column Footings. 
FIG. 4 • 


Tho wind moment M = W h, where W is the total wind force acting on the 
vertical projection of the tank and tower, and h is the height of tho center 
of gravity of this projected area above tho footing level on the ground. 

Tho stress in the coltimn will be v/cosoc,whore oc is the inclin¬ 

ation which the column makes with the vertical. 


151 







STRESSES IN LONG SPAN 3-HIHGED ARCH ROOF TRUSSES 


■t 


By David A. Molitor, C. E. 


Introductory » This subject is usually dismissed in our 
text books with a few remarks on wind forces and reaction con¬ 
ditions peculiar to the 3-hinged arch, followed by a statement 
that the stresses in the members may be found by any of the 
common methods, preferably by drawing a Maxwell stress diagram. 

For long span, shallow, trusses with many short web members, 
the Maxwell diagram is practically useless, and the method of 
moments involves many long lever arms with moment centers out¬ 
side the limits of the drawing, so that lengthy computations be¬ 
come necessary. 

The method presented herewith was evolved by the writer in 
an effort to obviate these difficulties thereby affording a 
simple process of stress analysis with some important checks. 

It consists (1) of finding the reactions and resultants of the 
external forces by well-known methods; r2)-then computing the 
chord stresses from the resultant polygon by the method of mom- 
ents; and (3) deriving the web stresses from the previously com¬ 
puted chord stresses by a new graphic method. 

The reactions and resultants are best found as follows: 

For diagonal wind loads , find the resultant of the external forces 
by means of a force and trial equilibrium polygon and then draw 
the final resultant polygon which gives the actual reactions and 
the successive resultants of the external forces from the crown 
hinge to the abutment hinge. The chord stresses may then be com¬ 
puted. For vertical Loads , compute the resultant of the external 
forces for the half arch, also the horizontal thrust and the 
vertical component of the end reaction, after which the final 
resultant polygon may be drawn directly without any trial equili¬ 
brium polygon. 

The following is a detailed description of the several steps 
comprising a complete stress analysis of the left half of a span 
•for diagonal wind loads as illustrated on the accompanying diagram. 

The Reactions . The wind loads wi to wg are combined into a 
force polygon, and a trial equilibrium polygon is drawn with 
an assumed pole _0' , thus locating the result^t W , and the straight 
closing line , A line through the pole 0’ , parallel to AC*, 
will divide the resultant A“C"° W into two reaction forces Pa and 
Pcf The resultant W must pass through the point A*, which is the 
point of intersection of the two end rays of the trial polygon. 

The resultant W is thus found in direction and magnitude, 

A final resultant polygon may now be drawn through the points 
A and C. To make this possible, the true pole 0 must first be 
located in the force polygon. 


t Publishec/ In Engineering News Apr! I 25, 1929, p 667. 


152 

















The direction of the pole £ must he parallel to the closing 
line and the reaction ^ acting at £, must pass through the 
right hand hinge of the ar^. Hence, draw C'*Q through 0“ and 
parallel to RoJ also draw £ from £, parallel to giving the 
intersection 0 as the true pole, ” 

Now draw the resultant polygon from A to jC, with the true 
pole 0, giving the actual end reactions R| and §2> which must 
intersect in a new point on the resultant W as a check. 

The position of any resultant of externally applied loads 
to the left of 0 is thus located for any section of the arch, and 
the magnitude oT any such resultant may be scaled from the force 
polygon. 

The chord Stresses can now be computed by the method of 
moments. Thus for member FG, the center of moments is n and the 
resultant in the bay £ is ^.7 kips, acting with a lever arm 
22,1 feet. Also, the lever arm from n to the member is 14.5 feet. 
Henoe the stress FG - 20,7 x • 3T.6 kips. Similarly all 

other chord stresses may be comptited. 

The Web Stresses are now derived from these chord stresses, 
obtaining one set of values from the bottom chord and a duplicate, 
or check, set of values from the top chord. 

.In the example here given, all the web stresses are found 
from the bot om chord stresses, while only a few are derived from 
the top chord to show how to insert the panel load W 3 ^ 

This method of determining the web stresses has the advantage 
of simplicity and accuracy, and affords a check not only for the 
web stresses but also for the previously computed chord stresses, 
which latter require no checking until a disparity in any one 
pair of results for a certain web member is discovered. Any 
erroneous chord stress is thus located. The method in its entirety 
is practically free from cumulative errors which are predominent in 
the ordinary stress diagram. 

To derive the web stresses from the computed bottom chord 
stresses, proceed as follows: From any point B' as a pole, draw 
a series of rays parallel to the botoom chord members, successively 
from A to £. When great accuracy is desired, this can be done 
by coordinate plotting, since for any large problem the coordinates 
of the panel points should always be computed before proceeding 
to a final stress analysis. 

The computed bottom chord stresses are now laid off on the 
respective rays B* D , B*K , B*N , etc, , using any convenient scale 
of forces, in this case the same scale as used in the force 
polygon. 


153 




o 

tf 

o 

X 

u 


±s 



154 















































The web stresses must now fit in between the points D, H, 

K, N, etc. , ao that all the members around a given panel point 
as n, will appear in a counter-clockwise order, thus , B*H , 

HG , and to form a closed polygon. In each such partial 
stress diagram all stresses are known except the two web stresses, 
which are thus determined in direction and amount. 

The direction of each stress is found by following around 
the partial stress diagram, starting with the two chord members 
of known direction. Sometimes this direction may be clockwise, 
as when the bottom chord members are in compression. In the case 
where some of the chord members of the same chord are in tension 
while others are in compression, it is necessary to draw the 
tension chords in a positive direction from the pole B' and the 
compression chord rays in the opposite direction. 

The stresses in the web members and have 

also been derived from the loaded top chord, merely as an 
Illustration, The order in which- the members 1^, FJ and the load 
W 3 all arranged is immaterial so long as the known quantities 
mre grouped together, permitting the two unknowns to close the 
polygon. The example shows a grouping which retains the system 
of rays from a common point as was done for the unloaded bottom 
chord. 

When dealing with vertical loads only, the horizontal thrust 
is computed and the final resultant polygon is drawn without the 
trial polygon. To compute the horizontal thrust,, take moments 
about C of 6l11 forces to the left of C, including the vertical 
reaction at A. The sura of these moments divided by the rise of 
the arch, gives the horizontal thrust. 


155 




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